 Time Flying  ### 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 = 1Output: 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 = 4Output: 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 = 2Output: 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}