Range Sum of BST — LeetCode 938 BST Pruning DFS
LeetCode 938 — Range Sum of BST, asked at Amazon, Facebook (Meta), Google and Apple. Use the BST property to prune entire out-of-range subtrees and run in O(h + k) time.
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LeetCode 938 — Range Sum of BST, asked at Amazon, Facebook (Meta), Google and Apple. Use the BST property to prune entire out-of-range subtrees and run in O(h + k) time.
LeetCode 700 — Search in a BST, asked at Amazon, Microsoft, Apple and Meta. Eliminate half the tree at each step using the BST property and finish in O(h) time, O(1) iterative space.
LC 450 Delete Node in a BST is a top FAANG interview problem testing all three deletion cases. Master the in-order successor strategy to pass BST questions at Amazon, Google, and Microsoft.
Solve LeetCode 538 Convert BST to Greater Tree using reverse in-order traversal asked at Google, Microsoft, and Amazon. Single O(n) pass with O(h) space.
Trim a BST so all values lie within [low, high] using recursive subtree pruning. LeetCode 669 is a Medium FAANG question asked at Amazon, Google, and Apple.
Encode a BST compactly using preorder traversal without null markers and rebuild it with min-max bounds. LeetCode 449 is a Medium FAANG question asked at Amazon, Google, and Meta.
Recover a BST where two nodes are swapped using in-order traversal to find the inversion pair. LeetCode 99 is asked at Amazon, Google, and Meta. Includes O(1) space Morris traversal solution.
LC 1373 Maximum Sum BST in Binary Tree finds the highest sum among all BST subtrees of a binary tree. The O(n) solution uses post-order DFS returning a 4-tuple of (is_bst, min, max, sum) metadata — a hard FAANG problem tested at Amazon and Google.
LC 1008 Construct BST from Preorder Traversal reconstructs a binary search tree in O(n) using min-max bounds to decide left vs right placement — a clean recursion problem tested at Amazon and Google that showcases BST property exploitation.
LeetCode 510 (Medium) frequently asked at Microsoft and Facebook. Find the inorder successor of a BST node when each node has a parent pointer, in O(h) time and O(1) extra space without access to the root.
LC 701 Insert into a BST traverses left or right based on value comparisons until finding a null position. The O(h) recursive solution is a BST fundamentals question tested at Amazon and Microsoft — always insert at a leaf.
LeetCode 897 (Easy). Rebuild a BST into a strictly right-skewed list using a single inorder traversal that rewires left and right pointers in place.