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Lowest Common Ancestor of a Binary Search Tree

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LeetCode 235. Lowest Common Ancestor of a Binary Search Tree

Given a binary search tree (BST), find the lowest common ancestor (LCA) node of two given nodes in the BST.

According to the definition of LCA on Wikipedia: "The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself)."

Examples

Example 1

Input:  root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8

6 (LCA) / \ 2 8 / \ / \ 0 4 7 9 / \ 3 5Output: 6 The LCA of nodes 2 and 8 is 6 (they are in different subtrees).

Example 2

Input:  root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2

The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself.

Example 3

Input:  root = [2,1], p = 2, q = 1
Output: 2

Constraints

  • The number of nodes in the tree is in the range [2, 10^5]
  • -10^9 <= Node.val <= 10^9
  • All Node.val are unique
  • p != q
  • p and q will exist in the BST

How the tests work: the tree is written as a level-order array and p/q as node values; your function receives the real TreeNode objects, and the answer is the value of the node you return.

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List of test cases (10)
Case 1
root = [6,2,8,0,4,7,9,null,null,3,5]
p = 2
q = 8
Case 2
root = [6,2,8,0,4,7,9,null,null,3,5]
p = 2
q = 4
Case 3
root = [2,1]
p = 2
q = 1
Case 4
root = [6,2,8,0,4,7,9,null,null,3,5]
p = 3
q = 5
Case 5
root = [6,2,8,0,4,7,9,null,null,3,5]
p = 7
q = 9
Case 6
root = [6,2,8,0,4,7,9,null,null,3,5]
p = 0
q = 5
Case 7
root = [6,2,8,0,4,7,9,null,null,3,5]
p = 3
q = 9
Case 8
root = [5,3,8,1,4,7,10]
p = 1
q = 4
Case 9
root = [5,3,8,1,4,7,10]
p = 7
q = 10
Case 10
root = [5,3,8,1,4,7,10]
p = 1
q = 10