LeetCode题解236.lowest-common-ancestor-of-a-binary-tree

题目地址

https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/

题目描述

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 tree:  root = [3,5,1,6,2,0,8,null,null,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 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.
 

Note:

All of the nodes' values will be unique.
p and q are different and both values will exist in the binary tree.

思路

这道题目是求解二叉树中，两个给定节点的最近的公共祖先。是一道非常经典的二叉树题目。

我们之前说过树是一种递归的数据结构，因此使用递归方法解决二叉树问题从写法上来看是最简单的，这道题目也不例外。

用递归的思路去思考树是一种非常重要的能力。

如果大家这样去思考的话，问题就会得到简化，我们的目标就是分别在左右子树进行查找p和q。 如果p没有在左子树，那么它一定在右子树（题目限定p一定在树中），
反之亦然。

对于具体的代码而言就是，我们假设这个树就一个结构，然后尝试去解决，然后在适当地方去递归自身即可。 如下图所示：

我们来看下核心代码：

  // 如果我们找到了p，直接进行返回，那如果下面就是q呢？ 其实这没有影响，但是还是要多考虑一下
  if (!root || root === p || root === q) return root;
  const left = lowestCommonAncestor(root.left, p, q); // 去左边找，我们期望返回找到的节点
  const right = lowestCommonAncestor(root.right, p, q);// 去右边找，我们期望返回找到的节点
  if (!left) return right; // 左子树找不到，返回右子树
  if (!right) return left; // 右子树找不到，返回左子树
  return root; // 左右子树分别有一个，则返回root

如果没有明白的话，请多花时间消化一下

关键点解析

• 用递归的思路去思考树

代码

代码支持： JavaScript， Python3

• JavaScript Code:

/**
 * 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) {
  if (!root || root === p || root === q) return root;
  const left = lowestCommonAncestor(root.left, p, q);
  const right = lowestCommonAncestor(root.right, p, q);
  if (!left) return right; // 左子树找不到，返回右子树
  if (!right) return left; // 右子树找不到，返回左子树
  return root; // 左右子树分别有一个，则返回root
};

• Python Code:

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if not root or root == p or root == q:
            return root
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)
        
        if not left:
            return right
        if not right:
            return left
        else:
            return root

扩展

如果递归的结束条件改为if (!root || root.left === p || root.right === q) return root; 代表的是什么意思，对结果有什么样的影响？

LeetCode题解104.maximum-depth-of-binary-tree

题目地址 https://leetcode.com/problems/maximum-depth-of-binary-tree/description/ 题目描述 Given a binary tree, find its maximum depth. The maximum depth is the number of nodes along the lo…

LeetCode题解124.binary-tree-maximum-path-sum

题目地址 https://leetcode.com/problems/binary-tree-maximum-path-sum/description/ 题目描述 Given a non-empty binary tree, find the maximum path sum. For this problem, a path is defined as a…

LeetCode题解1104.path-in-zigzag-labelled-binary-tree

题目地址 https://leetcode-cn.com/problems/path-in-zigzag-labelled-binary-tree/description/ 题目描述 在一棵无限的二叉树上，每个节点都有两个子节点，树中的节点 逐行 依次按 “之” 字形进行标记。 如下图所示，在奇数行（即，第一行、第三行、第五行……）中，按从左到右的顺序进行标…

LeetCode题解145.binary-tree-postorder-traversal

题目地址 https://leetcode.com/problems/binary-tree-postorder-traversal/description/ 题目描述 Given a binary tree, return the postorder traversal of its nodes' values. For example: Given bi…

LeetCode题解144.binary-tree-preorder-traversal

题目地址 https://leetcode.com/problems/binary-tree-preorder-traversal/description/ 题目描述 Given a binary tree, return the preorder traversal of its nodes' values. Example: Input: [1,null…