[leetcode]236. Lowest Common Ancestor of a Binary Tree二叉树最近公共祖先

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).”

Given the following binary search tree:  root = [3,5,1,6,2,0,8,null,null,7,4]

     _______3______
       /                  ___5__          ___1__
   /      \        /         6      _2       0       8
         /           7   4

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 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.

题意：

二叉树最近公共祖先

思路：

用自底向上(bottom-up)的思路，先看看是否能在root的左子树中找到p或q，再看看能否在右子树中找到，

• 如果两边都能找到，说明当前节点就是最近公共祖先
• 如果左边没找到，则说明p和q都在右子树
• 如果右边没找到，则说明p和q都在左子树

代码：

 1 class Solution {
 2       public TreeNode lowestCommonAncestor(TreeNode root, TreeNode node1, TreeNode node2) {
 3           if (root == null || root == node1 || root == node2) {
 4             return root;
 5         }
 6
 7         // Divide
 8         TreeNode left = lowestCommonAncestor(root.left, node1, node2);
 9         TreeNode right = lowestCommonAncestor(root.right, node1, node2);
10
11         // Conquer
12         if (left != null && right != null) {
13             return root;
14         }
15         if (left != null) {
16             return left;
17         }
18         if (right != null) {
19             return right;
20         }
21         return null;
22     }
23 }

原文地址：https://www.cnblogs.com/liuliu5151/p/9158486.html