BST — Advanced Patterns & Deep Dives

Pre-collect ALL nodes (eager):              Lazy traversal:
  O(n) space, O(n) init time                 O(h) space, O(1) amortized per call

  [1, 3, 5, 7, 9, ...]   ← full list        Stack: only current left spine
  load everything first                      push more only when needed

1. INIT:    Push entire left spine from root into stack
             (leftmost path = smallest values on top)

2. next():  Pop top node (= current smallest)
             → if it has a right child, push that subtree's left spine
             → return popped node's value

3. hasNext(): stack is non-empty

BST:          7
             / \
            3   15
               /  \
              9   20

Init: pushLeft(7) → stack = [7, 3]  (3 on top)

next() → pop 3, no right child → return 3,  stack = [7]
next() → pop 7, pushLeft(15)   → return 7,  stack = [15, 9]
next() → pop 9, no right child → return 9,  stack = [15]
next() → pop 15, pushLeft(20)  → return 15, stack = [20]
next() → pop 20, no right child → return 20, stack = []
hasNext() → false

class BSTIterator {
    private Stack<TreeNode> stack = new Stack<>();

    public BSTIterator(TreeNode root) {
        pushLeft(root);
    }

    private void pushLeft(TreeNode node) {
        while (node != null) {
            stack.push(node);
            node = node.left;
        }
    }

    public int next() {
        TreeNode node = stack.pop();
        if (node.right != null) {
            pushLeft(node.right);   // lazily expand right subtree
        }
        return node.val;
    }

    public boolean hasNext() {
        return !stack.isEmpty();
    }
}

class BSTIterator:
    def __init__(self, root):
        self.stack = []
        self._push_left(root)

    def _push_left(self, node):
        while node:
            self.stack.append(node)
            node = node.left

    def next(self) -> int:
        node = self.stack.pop()
        if node.right:
            self._push_left(node.right)
        return node.val

    def hasNext(self) -> bool:
        return bool(self.stack)

// java
// LC 1305 - All Elements in Two Binary Search Trees
// IDEA: TWO LAZY IN-ORDER ITERATORS + merge step
// time = O(m + n), space = O(h1 + h2) excluding the output list
public List<Integer> getAllElements(TreeNode root1, TreeNode root2) {
    Deque<TreeNode> s1 = new ArrayDeque<>(), s2 = new ArrayDeque<>();
    List<Integer> res = new ArrayList<>();
    pushLeft(s1, root1);
    pushLeft(s2, root2);

    while (!s1.isEmpty() || !s2.isEmpty()) {
        // take from whichever iterator currently exposes the smaller value
        Deque<TreeNode> pick;
        if (s2.isEmpty() || (!s1.isEmpty() && s1.peek().val <= s2.peek().val)) {
            pick = s1;
        } else {
            pick = s2;
        }
        TreeNode node = pick.pop();
        pushLeft(pick, node.right);   // lazily expand only the stack we consumed
        res.add(node.val);
    }
    return res;
}

private void pushLeft(Deque<TreeNode> st, TreeNode node) {
    while (node != null) { st.push(node); node = node.left; }
}

# python
# LC 1305 - All Elements in Two Binary Search Trees
# IDEA: TWO LAZY IN-ORDER ITERATORS + merge step
# time = O(m + n), space = O(h1 + h2) excluding the output list
class Solution(object):
    def getAllElements(self, root1, root2):
        st1, st2, res = [], [], []

        def push_left(st, node):
            while node:
                st.append(node)
                node = node.left

        push_left(st1, root1)
        push_left(st2, root2)

        while st1 or st2:
            # pop from the iterator exposing the smaller current value
            if not st2 or (st1 and st1[-1].val <= st2[-1].val):
                node = st1.pop()
                push_left(st1, node.right)
            else:
                node = st2.pop()
                push_left(st2, node.right)
            res.append(node.val)

        return res

1. Push LEFT SPINE only — this gives smallest-first access
2. On pop: expand right subtree's left spine lazily
3. Stack depth = O(h), not O(n) — critical for tall/large trees
4. Amortized O(1) per next(): each node is pushed and popped exactly once
5. Enables partial traversal — stop early without wasting work

1. Walk in-order, keeping ONE extra pointer: `prev` (the previously visited node)
2. Every time prev.val > cur.val  -> that's a DROP
      first  = prev   (only on the FIRST drop)   <- the "too large" node
      second = cur    (on EVERY drop)            <- the "too small" node
3. Swap first.val <-> second.val

Core Insight:
  Valid BST in-order traversal → strictly INCREASING sequence

  Example valid BST:        In-order: [1, 2, 3, 4, 5, 6, 7]
         4                            ↑  ↑  ↑  ↑  ↑  ↑  ↑
        / \                          strictly increasing ✓
       2   6
      / \ / \
     1  3 5  7

  If two nodes SWAPPED:     In-order: [1, 6, 3, 4, 5, 2, 7]
         4                                 ↓        ↓
        / \                            DROP here  DROP here
       6   2    (swapped!)                 ↓        ↓
      / \ / \                         first=6   second=2
     1  3 5  7

Finding "drops" in sequence = finding swapped nodes!

Case 1: ADJACENT nodes swapped (1 drop)
  Valid:   [1, 2, 3, 4, 5]
  Swapped: [1, 3, 2, 4, 5]  ← swap 2 and 3
                ↓
           one drop: 3 > 2
           first = 3 (prev at drop)
           second = 2 (current at drop)

Case 2: DISTANT nodes swapped (2 drops)
  Valid:   [1, 2, 3, 4, 5, 6, 7]
  Swapped: [1, 6, 3, 4, 5, 2, 7]  ← swap 2 and 6
                ↓           ↓
           drop 1: 6 > 3    drop 2: 5 > 2
           first = 6        second = 2 (update!)

Key: first is set at FIRST drop, second is ALWAYS updated at each drop

# python
# LC 99 - Recover Binary Search Tree
# IDEA: IN-ORDER DFS + BST PROPERTY (in-order of a valid BST is strictly increasing)
# time = O(n), space = O(h) recursion stack
class Solution(object):
    def recoverTree(self, root):
        self.first = None    # the "too large" node  (prev at the 1st drop)
        self.second = None   # the "too small" node  (cur  at the last drop)
        self.prev = None     # previously visited node in in-order order

        self.helper(root)

        # NOTE !!! swap VALUES, never rewire the nodes
        if self.first and self.second:
            self.first.val, self.second.val = self.second.val, self.first.val

    def helper(self, node):
        if not node:
            return

        # 1) left
        self.helper(node.left)

        # 2) current : compare against the previous in-order node
        #    NOTE !!! `self.prev` guard — prev is None only for the leftmost node
        if self.prev and self.prev.val > node.val:
            if self.first is None:
                self.first = self.prev   # FIRST drop only
            self.second = node           # EVERY drop (overwrites on the 2nd)

        self.prev = node                 # current becomes previous

        # 3) right
        self.helper(node.right)

// java
// LC 99 Recover Binary Search Tree
// Time: O(N), Space: O(H)

/**
 * Key variables:
 * - first:  The first node where order is violated (the larger one in first drop)
 * - second: The second node where order is violated (the smaller one in last drop)
 * - prev:   The previously visited node in in-order traversal
 */
private TreeNode first = null;
private TreeNode second = null;
// NOTE: keep prev = null and guard with `prev != null`.
// A `new TreeNode(Integer.MIN_VALUE)` sentinel is NOT safe here — LC 99 allows
// node values down to -2^31, so a real node can equal the sentinel.
private TreeNode prev = null;

public void recoverTree(TreeNode root) {
    if (root == null) return;

    // Step 1: In-order traversal to find the two swapped nodes
    inorder(root);

    // Step 2: Swap the values (not the nodes themselves!)
    if (first != null && second != null) {
        int temp = first.val;
        first.val = second.val;
        second.val = temp;
    }
}

private void inorder(TreeNode root) {
    if (root == null) return;

    // 1. Go Left
    inorder(root.left);

    // 2. Process current node - detect violation
    if (prev != null && root.val < prev.val) {
        // Found a DROP in the sequence!
        if (first == null) {
            first = prev;    // First drop: prev is the "too large" node
        }
        second = root;       // Always update: current is the "too small" node
    }

    // Update prev for next comparison
    prev = root;

    // 3. Go Right
    inorder(root.right);
}

# python
# LC 99 - Recover BST, O(1) extra space
# IDEA: MORRIS IN-ORDER + the same (first / second / prev) drop detection
# time = O(n) (each edge traversed at most 3x), space = O(1)
class Solution(object):
    def recoverTree(self, root):
        first = second = prev = None
        cur = root

        while cur:
            if not cur.left:
                # no left subtree -> visit cur, then go right
                if prev and prev.val > cur.val:
                    if not first:
                        first = prev
                    second = cur
                prev = cur
                cur = cur.right
            else:
                # find in-order predecessor of cur
                pred = cur.left
                while pred.right and pred.right is not cur:
                    pred = pred.right

                if not pred.right:
                    pred.right = cur      # thread it, descend left
                    cur = cur.left
                else:
                    pred.right = None     # NOTE !!! un-thread (restore structure)
                    if prev and prev.val > cur.val:
                        if not first:
                            first = prev
                        second = cur
                    prev = cur
                    cur = cur.right

        if first and second:
            first.val, second.val = second.val, first.val

The algorithm handles BOTH cases with ONE logic:

1. "first" is set only ONCE at the first drop
   → This captures the "too large" node that was swapped

2. "second" is ALWAYS updated at every drop
   → For adjacent swap: only 1 drop, second = the "too small" node ✓
   → For distant swap: 2 drops, second gets overwritten to correct node ✓

Example (distant swap: 2 and 6):
  Sequence: [1, 6, 3, 4, 5, 2, 7]

  At index 2 (val=3): prev=6, curr=3, 6 > 3 → DROP!
    first = 6 (set once)
    second = 3

  At index 5 (val=2): prev=5, curr=2, 5 > 2 → DROP!
    first = 6 (unchanged)
    second = 2 (updated!) ← This is the correct second node

  Swap 6 and 2 → BST recovered ✓

Why swap VALUES, not NODE OBJECTS?

  Swapping node objects means rewiring parent/child pointers for BOTH nodes,
  plus handling edge cases (root, adjacent nodes, left/right children).
  This is extremely complex and error-prone.

  Swapping values is O(1) and preserves the tree structure:
  - All parent→child links stay the same
  - All left/right subtree relationships stay the same
  - Only the .val fields change — the BST property is restored

  Example:
       4                    4
      / \    swap vals      / \
     6   2   of 6 & 2 →   2   6    ← valid BST!
    / \ / \               / \ / \
   1  3 5  7             1  3 5  7

  The nodes stay in the SAME positions. Only their values change.
  Tree structure (edges) is completely untouched.

// BAD: Swapping node references does NOTHING to the tree
TreeNode temp = first;
first = second;       // Only swaps local variable pointers!
second = temp;        // The actual tree is unchanged.

// GOOD: Swap the values stored inside the nodes
int temp = first.val;
first.val = second.val;
second.val = temp;

// BAD: Stops after first drop
if (prev.val > root.val) {
    first = prev;
    second = root;
    return;  // Wrong! Miss the second drop for distant swaps
}

// GOOD: Always update second, handles both cases
if (prev.val > root.val) {
    if (first == null) first = prev;
    second = root;  // Always update!
}

// BAD: LC 99 allows node values down to -2^31, so a real node can EQUAL
//      the sentinel — and a float('-inf') sentinel in Python isn't a TreeNode.
TreeNode prev = new TreeNode(Integer.MIN_VALUE);

// GOOD: prev = null, and guard the comparison. null happens exactly once
//       (the leftmost node), where there is nothing to compare against anyway.
TreeNode prev = null;
...
if (prev != null && prev.val > root.val) { ... }

# BAD: overwrites the correct answer on a distant swap
if prev.val > node.val:
    first = prev      # gets clobbered at the 2nd drop -> wrong pair
    second = node

# GOOD: first is write-once, second is write-always
if prev.val > node.val:
    if first is None:
        first = prev
    second = node

1. In-order traversal of valid BST = STRICTLY INCREASING sequence
   → This is the MOST IMPORTANT property for BST problems

2. Detecting violations = finding "drops" in the sequence
   → prev.val > curr.val means we found a violation

3. Two-drop pattern handles both adjacent and distant swaps
   → first: set at FIRST drop (the larger misplaced node)
   → second: ALWAYS update (the smaller misplaced node)

4. Swap VALUES, not nodes
   → Much simpler, preserves tree structure

Plain BST answers "is x present?".
Augmented BST answers "what is the kth largest?" — because each node knows
how many values live beneath it.

           node
          /    \
      left      right
    (a nodes) (b nodes)

  The b values in `right` are ALL larger than node.val.
  So, ranking from the largest downwards:

    ranks 1 .. b        → live in the right subtree
    rank  b + 1         → is node itself
    ranks b + 2 ..      → live in the left subtree

  kthLargest(node, k):
      b = size(node.right)
      if k <= b       → recurse right, k unchanged
      if k == b + 1   → return node.val
      else            → recurse left with k - (b + 1)
                         (we just skipped the right subtree AND node)

BST after [4,5,8,2] then inserting 3   (count in parentheses)

              4(5)
             /    \
          2(2)     5(2)
             \        \
             3(1)      8(1)

kthLargest(root=4, k=3):
  b = size(node.right = 5) = 2      → subtree {5, 8} holds the 1st and 2nd largest
  k = 3 == b + 1                    → node 4 IS the 3rd largest → return 4   ✓
  (sorted values {2,3,4,5,8}; 3rd largest = 4)

add(5) next → tree grows to {2,3,4,5,5,8}, kthLargest(root, 3) = 5

// java
// LC 703 - Kth Largest Element in a Stream
// IDEA: ORDER-STATISTIC BST — augment each node with subtree size, then rank-descend
// time = O(h) per add (O(log n) balanced, O(n) skewed), space = O(n)
class KthLargest {

    private static class Node {
        int val;
        int count = 1;          // number of nodes in the subtree rooted here
        Node left, right;
        Node(int v) { val = v; }
    }

    private final int k;
    private Node root;

    public KthLargest(int k, int[] nums) {
        this.k = k;
        for (int v : nums) root = insert(root, v);
    }

    private Node insert(Node node, int val) {
        if (node == null) return new Node(val);
        node.count++;                                    // one more node below/at this root
        if (val < node.val) node.left = insert(node.left, val);
        else node.right = insert(node.right, val);       // duplicates go right (multiset)
        return node;
    }

    private int size(Node node) { return node == null ? 0 : node.count; }

    private int kthLargest(Node node, int k) {
        int rightSize = size(node.right);                // count of values > node.val
        if (k <= rightSize) return kthLargest(node.right, k);
        if (k == rightSize + 1) return node.val;         // node itself is the answer
        return kthLargest(node.left, k - rightSize - 1); // skipped right subtree + node
    }

    public int add(int val) {
        root = insert(root, val);
        return kthLargest(root, k);
    }
}

# python
# LC 703 - Kth Largest Element in a Stream
# IDEA: ORDER-STATISTIC BST — augment each node with subtree size, then rank-descend
# time = O(h) per add (O(log n) balanced, O(n) skewed), space = O(n)
class CountNode(object):
    def __init__(self, val):
        self.val = val
        self.left = None
        self.right = None
        self.count = 1          # size of the subtree rooted here


class KthLargest(object):
    def __init__(self, k, nums):
        self.k = k
        self.root = None
        for v in nums:
            self.root = self._insert(self.root, v)

    def _insert(self, node, val):
        if not node:
            return CountNode(val)
        node.count += 1                                   # subtree grew by 1
        if val < node.val:
            node.left = self._insert(node.left, val)
        else:
            node.right = self._insert(node.right, val)    # duplicates go right
        return node

    def _size(self, node):
        return node.count if node else 0

    def _kth_largest(self, node, k):
        right_size = self._size(node.right)               # values greater than node.val
        if k <= right_size:
            return self._kth_largest(node.right, k)
        if k == right_size + 1:
            return node.val                               # node itself
        return self._kth_largest(node.left, k - right_size - 1)

    def add(self, val):
        self.root = self._insert(self.root, val)
        return self._kth_largest(self.root, self.k)

With size-augmented nodes, all of these are O(h) descents:

  kthLargest(k)        → shown above
  kthSmallest(k)       → mirror it: use size(node.left)
  rank(x)              → count of values < x: add size(left)+1 each time you go right
  countInRange(lo,hi)  → rank(hi) - rank(lo)
  median              → kthSmallest(n/2) using size(root)

  ⚠️ The ONLY maintenance cost: every insert/delete must fix `count`
     on every node along the path. Forgetting this silently corrupts all ranks.

1. A bare BST answers membership; a SIZE-AUGMENTED BST answers RANK
2. kth largest = descend using size(right): k<=b → right, k==b+1 → node, else left with k-b-1
3. Bump `count` on the way DOWN during insert — every node on the path
4. Duplicates: send them consistently to one side (right here) → BST becomes a multiset
5. State the O(n) skewed worst case and name TreeMap / Red-Black as the fix

This is the standard answer to "what if the data keeps changing?" follow-ups
on LC 230 and any kth-element question.

def sorted_list_to_bst(head):
    """
    Convert sorted linked list to balanced BST
    Two approaches: 1) Two pointers to find middle
                   2) Inorder simulation
    Time: O(n log n) or O(n), Space: O(log n)
    """
    # Approach 1: Find middle with slow-fast pointers
    def find_middle(left, right):
        slow = fast = left
        while fast != right and fast.next != right:
            slow = slow.next
            fast = fast.next.next
        return slow

    def convert(left, right):
        if left == right:
            return None

        mid = find_middle(left, right)
        root = TreeNode(mid.val)
        root.left = convert(left, mid)
        root.right = convert(mid.next, right)
        return root

    return convert(head, None)

def bst_from_preorder(preorder):
    """
    Construct BST from preorder traversal
    Use BST property: left < root < right
    Time: O(n), Space: O(h)
    """
    def build(min_val, max_val):
        nonlocal idx
        if idx >= len(preorder):
            return None

        val = preorder[idx]
        if val < min_val or val > max_val:
            return None

        idx += 1
        root = TreeNode(val)
        root.left = build(min_val, val)
        root.right = build(val, max_val)
        return root

    idx = 0
    return build(float('-inf'), float('inf'))

   in-order DFS                      mid-as-root recursion
  ────────────────►                  ─────────────────────►
  BST  ──────────► sorted array ───► balanced BST
  (any shape)      [1,2,3,4]         (height = ceil(log2(n+1)))
   any BST flattens to               any sorted array rebuilds
   the SAME sorted order             to a BALANCED BST

build(l, r) puts nodes[mid] on top, with
   left  subtree = build(l, mid-1)   -> size = mid - l
   right subtree = build(mid+1, r)   -> size = r - mid
   |left size - right size| <= 1     at EVERY node
=> heights differ by at most 1       at EVERY node   ← exactly the problem's definition
=> total height = floor(log2(n)) + 1 = ceil(log2(n+1))

# python
# LC 1382 - Balance a Binary Search Tree
# IDEA: IN-ORDER DFS (BST -> sorted array) + mid-as-root rebuild (LC 108)
# time  = O(n)  -- n for the traversal, n for the rebuild
# space = O(n)  -- the array; + O(h_in) recursion for in-order, O(log n) for the build
class Solution(object):
    def balanceBST(self, root):
        if not root:
            return None

        self.arr = []
        self.in_order(root)                              # step 1: BST -> sorted
        return self.build(0, len(self.arr) - 1)          # step 2: sorted -> balanced BST

    # NOTE !!! in-order, because ONLY in-order gives sorted output for a BST
    def in_order(self, node):
        if not node:
            return
        self.in_order(node.left)
        self.arr.append(node.val)
        self.in_order(node.right)

    # NOTE !!! index bounds, NOT slicing -> avoids O(n log n) copying
    def build(self, l, r):
        # base case: empty range -> None (this also CLEARS stale child pointers)
        if l > r:
            return None

        mid = l + (r - l) // 2                           # middle becomes root
        node = TreeNode(self.arr[mid])
        node.left = self.build(l, mid - 1)               # left half
        node.right = self.build(mid + 1, r)              # right half
        return node

# python
# LC 1382 - node-reuse variant (no new allocations)
# time = O(n), space = O(n) for the node list
class Solution(object):
    def balanceBST(self, root):
        nodes = []

        def in_order(node):
            if not node:
                return
            in_order(node.left)
            nodes.append(node)                # collect NODES, not values
            in_order(node.right)

        def build(l, r):
            if l > r:
                return None
            mid = l + (r - l) // 2
            node = nodes[mid]
            node.left = build(l, mid - 1)      # MUST assign both children,
            node.right = build(mid + 1, r)     # else old pointers linger
            return node

        in_order(root)
        return build(0, len(nodes) - 1)

// java
// LC 1382 - Balance a Binary Search Tree
// IDEA: in-order into a list, then rebuild with mid-as-root
// time = O(n), space = O(n)
public TreeNode balanceBST(TreeNode root) {
    List<Integer> sorted = new ArrayList<>();
    inorder(root, sorted);
    return build(sorted, 0, sorted.size() - 1);
}

private void inorder(TreeNode node, List<Integer> out) {
    if (node == null) return;
    inorder(node.left, out);
    out.add(node.val);           // in-order => ascending
    inorder(node.right, out);
}

private TreeNode build(List<Integer> a, int l, int r) {
    if (l > r) return null;
    int mid = l + (r - l) / 2;   // NOTE: not (l + r) / 2 -> avoids int overflow
    TreeNode node = new TreeNode(a.get(mid));
    node.left  = build(a, l, mid - 1);
    node.right = build(a, mid + 1, r);
    return node;
}

 step 1: in-order          step 2: build(0, 3), mid = 0 + (3-0)//2 = 1
   1                          arr = [1, 2, 3, 4]
    \                                    ^ mid
     2       in-order                            2                h = 3
      \      ────────►  [1, 2, 3, 4]           /   \
       3                                      1     3      build(0,0) | build(2,3)
        \                                            \
         4                                            4     build(3,3), mid=3

# ❌ Pre-order / post-order -> NOT sorted -> the rebuilt tree isn't a BST at all
# ❌ Slicing: build(arr[:mid]) + build(arr[mid+1:])  -> O(n log n) copying (works, but wasteful)
# ❌ Reusing nodes with only `if l <= mid-1: node.left = ...`  -> stale pointers survive
# ❌ mid = (l + r) / 2 in Java -> int overflow on huge ranges; use l + (r - l) / 2
# ❌ Recursive in-order on a FULLY SKEWED BST with n = 10^4 (LC 1382's max):
#      Python default recursionlimit is 1000 -> RecursionError.
#      Fix: iterative in-order (stack), or sys.setrecursionlimit(10**5)

# ✅ iterative in-order — immune to the skewed-tree recursion blowup
def in_order_iter(root):
    out, stack, cur = [], [], root
    while cur or stack:
        while cur:
            stack.append(cur)
            cur = cur.left
        cur = stack.pop()
        out.append(cur.val)
        cur = cur.right
    return out

1. VINE     : right-rotate away every left child  -> a right-leaning linked list
              (root)-> 1 -> 2 -> 3 -> 4 -> ...
2. COUNT    : walk the vine to get n
3. COMPRESS : m = 2^floor(log2(n+1)) - 1          (size of the largest perfect tree)
              make_rotations(n - m)               (level the excess leaves)
              while m > 1: m //= 2; make_rotations(m)   (halve repeatedly)

def generate_trees(n):
    """
    Generate all structurally unique BSTs with n nodes
    Catalan number of trees: C(n) = (2n)! / ((n+1)! * n!)
    Time: O(4^n / n^(3/2)), Space: O(4^n / n^(3/2))
    """
    if n == 0:
        return []

    def generate(start, end):
        if start > end:
            return [None]

        all_trees = []
        for root_val in range(start, end + 1):
            left_trees = generate(start, root_val - 1)
            right_trees = generate(root_val + 1, end)

            for left in left_trees:
                for right in right_trees:
                    root = TreeNode(root_val)
                    root.left = left
                    root.right = right
                    all_trees.append(root)

        return all_trees

    return generate(1, n)

// Java — Plain Recursion (LC 95)
public List<TreeNode> generateTrees(int n) {
    if (n == 0) return new ArrayList<>();
    return buildTrees(1, n);
}

private List<TreeNode> buildTrees(int start, int end) {
    List<TreeNode> allTrees = new ArrayList<>();
    if (start > end) {
        allTrees.add(null);  // important: null represents empty subtree
        return allTrees;
    }
    for (int i = start; i <= end; i++) {
        // left subtree candidates: [start, i-1]
        List<TreeNode> leftSubtrees = buildTrees(start, i - 1);
        // right subtree candidates: [i+1, end]
        List<TreeNode> rightSubtrees = buildTrees(i + 1, end);
        // Cartesian product: connect each left × right to root i
        for (TreeNode left : leftSubtrees) {
            for (TreeNode right : rightSubtrees) {
                TreeNode root = new TreeNode(i);
                root.left = left;
                root.right = right;
                allTrees.add(root);
            }
        }
    }
    return allTrees;
}

// Java — Memoized Recursion (LC 95)
public List<TreeNode> generateTrees_memo(int n) {
    Map<Pair<Integer, Integer>, List<TreeNode>> memo = new HashMap<>();
    return allPossibleBST(1, n, memo);
}

private List<TreeNode> allPossibleBST(int start, int end,
        Map<Pair<Integer, Integer>, List<TreeNode>> memo) {
    List<TreeNode> res = new ArrayList<>();
    if (start > end) { res.add(null); return res; }
    if (memo.containsKey(new Pair<>(start, end)))
        return memo.get(new Pair<>(start, end));
    for (int i = start; i <= end; ++i) {
        List<TreeNode> leftSubs = allPossibleBST(start, i - 1, memo);
        List<TreeNode> rightSubs = allPossibleBST(i + 1, end, memo);
        for (TreeNode left : leftSubs)
            for (TreeNode right : rightSubs)
                res.add(new TreeNode(i, left, right));
    }
    memo.put(new Pair<>(start, end), res);
    return res;
}

def num_trees(n):
    """
    Count number of structurally unique BSTs with n nodes
    Uses Catalan Number: C(n) = C(0)*C(n-1) + C(1)*C(n-2) + ... + C(n-1)*C(0)
    Time: O(n^2), Space: O(n)
    """
    dp = [0] * (n + 1)
    dp[0] = dp[1] = 1

    for nodes in range(2, n + 1):
        for root in range(1, nodes + 1):
            left = root - 1
            right = nodes - root
            dp[nodes] += dp[left] * dp[right]

    return dp[n]

// Pattern 6.3: From Preorder (LC 1008)
private int idx = 0;

public TreeNode bstFromPreorder(int[] preorder) {
    return build(preorder, Integer.MIN_VALUE, Integer.MAX_VALUE);
}

private TreeNode build(int[] preorder, int min, int max) {
    if (idx >= preorder.length) return null;

    int val = preorder[idx];
    if (val < min || val > max) return null;

    idx++;
    TreeNode root = new TreeNode(val);
    root.left = build(preorder, min, val);
    root.right = build(preorder, val, max);
    return root;
}

# BAD: Creates new arrays O(n) space each recursion
def build(nums):
    mid = len(nums) // 2
    root.left = build(nums[:mid])  # O(n) space

# GOOD: Use indices
def build(left, right):
    mid = (left + right) // 2
    root.left = build(left, mid - 1)

# BAD: Wrong boundary
mid = len(nums) // 2
root.left = build(nums[:mid-1])  # Skips element!

# GOOD: Correct boundaries
root.left = build(nums[:mid])  # Includes all left elements

# BAD: No bounds checking
def build():
    val = preorder[idx]
    root = TreeNode(val)  # May violate BST property!

# GOOD: Check min/max bounds
def build(min_val, max_val):
    if val < min_val or val > max_val:
        return None

# BAD: Wrong combination
for i in range(1, n+1):
    dp[n] += dp[i] + dp[n-i]  # Should multiply!

# GOOD: Multiply left and right counts
for i in range(1, n+1):
    dp[n] += dp[i-1] * dp[n-i]

# BAD: Can overflow in Java/C++
mid = (left + right) / 2

# GOOD: Prevent overflow
mid = left + (right - left) // 2

// java
// LC 1110 - Delete Nodes And Return Forest
// IDEA: DFS + "return null to detach" + pass `isRoot` down
// time = O(n), space = O(h + d)   d = |to_delete|
// NOTE: keep the state LOCAL — instance fields would leak between calls
//       (LeetCode reuses one Solution object for every test case)
public List<TreeNode> delNodes(TreeNode root, int[] to_delete) {
    Set<Integer> toDelete = new HashSet<>();
    for (int v : to_delete) toDelete.add(v);
    List<TreeNode> forest = new ArrayList<>();
    walk(root, true, toDelete, forest);
    return forest;
}

private TreeNode walk(TreeNode node, boolean isRoot, Set<Integer> toDelete, List<TreeNode> forest) {
    if (node == null) return null;
    boolean deleted = toDelete.contains(node.val);

    // a surviving node whose parent vanished starts a new tree
    if (isRoot && !deleted) forest.add(node);

    // children are "roots" iff THIS node is being deleted
    node.left  = walk(node.left,  deleted, toDelete, forest);
    node.right = walk(node.right, deleted, toDelete, forest);

    // returning null is what actually detaches this node from its parent
    return deleted ? null : node;
}

# python
# LC 1110 - Delete Nodes And Return Forest
# IDEA: DFS + "return None to detach" + pass `is_root` down
# time = O(n), space = O(h + d)
class Solution(object):
    def delNodes(self, root, to_delete):
        to_del = set(to_delete)
        forest = []

        def dfs(node, is_root):
            if not node:
                return None
            deleted = node.val in to_del

            # surviving node with a deleted parent → new forest root
            if is_root and not deleted:
                forest.append(node)

            # children become roots only if THIS node is deleted
            node.left = dfs(node.left, deleted)
            node.right = dfs(node.right, deleted)

            return None if deleted else node

        dfs(root, True)
        return forest

# python code

# LC 450 Delete Node in a BST
# IDEA : RECURSION + BST PROPERTY
#### 2 CASES :
#   -> CASE 1 : root.val == key and NO right subtree 
#                -> swap root and root.left, return root.left
#   -> CASE 2 : root.val == key and THERE IS right subtree
#                -> 1) go to 1st RIGHT sub tree
#                -> 2) iterate to deepest LEFT subtree
#                -> 3) swap root and  `deepest LEFT subtree` then return root
class Solution(object):
    def deleteNode(self, root, key):
        if not root: return None
        if root.val == key:
            # case 1 : NO right subtree 
            if not root.right:
                left = root.left
                return left
            # case 2 : THERE IS right subtree
            else:
                ### NOTE : find min in "right" sub-tree
                #           -> because BST property, we ONLY go to 1st right tree (make sure we find the min of right sub-tree)
                #           -> then go to deepest left sub-tree
                right = root.right
                while right.left:
                    right = right.left
                root.val, right.val = right.val, root.val
        root.left = self.deleteNode(root.left, key)
        root.right = self.deleteNode(root.right, key)
        return root

// java
// LC 1026 - Maximum Difference Between Node and Ancestor
// IDEA: BOUNDS PROPAGATION (same skeleton as LC 98, but maximize instead of reject)
// time = O(n), space = O(h)
public int maxAncestorDiff(TreeNode root) {
    if (root == null) return 0;
    return dfs(root, root.val, root.val);
}

private int dfs(TreeNode node, int mn, int mx) {
    // at a leaf's child, the widest ancestor spread on this path is mx - mn
    if (node == null) return mx - mn;

    mn = Math.min(mn, node.val);
    mx = Math.max(mx, node.val);

    return Math.max(dfs(node.left, mn, mx), dfs(node.right, mn, mx));
}

# python
# LC 1026 - Maximum Difference Between Node and Ancestor
# IDEA: BOUNDS PROPAGATION (same skeleton as LC 98, but maximize instead of reject)
# time = O(n), space = O(h)
class Solution(object):
    def maxAncestorDiff(self, root):
        def dfs(node, mn, mx):
            if not node:
                return mx - mn          # widest spread seen on this path
            mn = min(mn, node.val)
            mx = max(mx, node.val)
            return max(dfs(node.left, mn, mx), dfs(node.right, mn, mx))

        if not root:
            return 0
        return dfs(root, root.val, root.val)

Key idea: |a - b| over an ancestor/descendant pair on a path is maximized by
          the path's MIN and MAX — so you never need to compare all pairs,
          only carry two numbers down.

Validate BST (98)          Max ancestor diff (1026)
  bounds = legal window      bounds = running min/max of ancestors
  node outside → return F    node outside → widen the window
  combine with AND           combine with MAX