[LeetCode] 236. Lowest Common Ancestor of a Binary 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:

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:

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.

二叉树的最近公共祖先。

题目跟235题很接近，唯一的不同是235题是一个二叉搜索树（BST），本题是一个二叉树。235题可以用递归做，但是本题比较难，需要用到后序遍历（postorder），因为无法用BST的性质判断p和q到底是在左子树还是在右子树。注意需要判断一个corner case，如果根节点为空，或者p和q里面有任何一个节点的 val 和根节点的 val 一样的情况，此时需要返回 root。

时间O(n)

空间O(h) - 树的高度

Java实现

 1 class Solution {
 2     public TreeNode lowestCommonAncestor(TreeNode root, TreeNode p, TreeNode q) {
 3         if (root == null || root == p || root == q) return root;
 4         TreeNode left = lowestCommonAncestor(root.left, p, q);
 5         TreeNode right = lowestCommonAncestor(root.right, p, q);
 6         if (left != null && right != null) {
 7             return root;
 8         }
 9         return left == null ? right : left;
10     }
11 }

JavaScript实现

 1 /**
 2  * @param {TreeNode} root
 3  * @param {TreeNode} p
 4  * @param {TreeNode} q
 5  * @return {TreeNode}
 6  */
 7 var lowestCommonAncestor = function (root, p, q) {
 8     if (root == null || root == p || root == q) return root;
 9     let left = lowestCommonAncestor(root.left, p, q);
10     let right = lowestCommonAncestor(root.right, p, q);
11     if (left !== null && right !== null) {
12         return root;
13     }
14     return left == null ? right : left;
15 };

根据代码，跑一下example 1，p = 5，q = 1。因为是后序遍历所以先从叶子节点开始找，节点7和节点4满足11行（JS）所以会return他们本身；再往上找的时候会找到他们的父节点2，再找到6和2的父节点5。此时5和1满足11行，所以会return根节点3。