 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 = 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
}``````