236.Lowest Common Ancestor of a Binary Search Tree

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

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).”

Example 1:

3
     /   \
    5      1
   / \    / \
  6   2  0   8
     / \
    7   4

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3

Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5

Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Example 3:

1
  /
 2

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

Constraints:

The number of nodes in the tree is in the range [2, 105]. -109 <= Node.val <= 109 All Node.val are unique. p != q p and q will exist in the tree.

Analyze

• Using post-order traversal of birnary tree in recursive.

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {TreeNode}
 */
var lowestCommonAncestor = function(root, p, q) {
  return lca(root, p, q)
};

var lca = function(node, p, q) {
  if (!node) return null
  // node is the lowest common ancestor of p and q.
  if (node.val === p.val || node.val === q.val) return node

  const leftLca = lca(node.left, p, q)
  const rightLca = lca(node.right, p, q)

  // get value of post-order traversal
  if (!leftLca) {
    return rightLca
  }
  if (!rightLca) {
    return leftLca
  }
  return node
}