130 lines
4.2 KiB
Markdown
130 lines
4.2 KiB
Markdown
## 题目地址
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https://leetcode.com/problems/lowest-common-ancestor-of-a-binary-tree/description/
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## 题目描述
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```
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Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.
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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).”
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Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]
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```
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```
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Example 1:
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Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
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Output: 3
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Explanation: The LCA of nodes 5 and 1 is 3.
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Example 2:
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Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
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Output: 5
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Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.
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Note:
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All of the nodes' values will be unique.
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p and q are different and both values will exist in the binary tree.
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```
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## 思路
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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。
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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。
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用递归的思路去思考树是一种非常重要的能力。
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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），
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反之亦然。
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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：
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我们来看下核心代码：
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```js
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// 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
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if (!root || root === p || root === q) return root;
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const left = lowestCommonAncestor(root.left, p, q); // 去左边找，我们期望返回找到的节点
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const right = lowestCommonAncestor(root.right, p, q);// 去右边找，我们期望返回找到的节点
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if (!left) return right; // 左子树找不到，返回右子树
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if (!right) return left; // 右子树找不到，返回左子树
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return root; // 左右子树分别有一个，则返回root
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```
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> 如果没有明白的话，请多花时间消化一下
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## 关键点解析
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- 用递归的思路去思考树
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## 代码
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代码支持： JavaScript， Python3
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- JavaScript Code:
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```js
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/**
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* Definition for a binary tree node.
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* function TreeNode(val) {
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* this.val = val;
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* this.left = this.right = null;
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* }
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*/
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/**
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* @param {TreeNode} root
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* @param {TreeNode} p
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* @param {TreeNode} q
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* @return {TreeNode}
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*/
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var lowestCommonAncestor = function(root, p, q) {
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if (!root || root === p || root === q) return root;
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const left = lowestCommonAncestor(root.left, p, q);
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const right = lowestCommonAncestor(root.right, p, q);
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if (!left) return right; // 左子树找不到，返回右子树
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if (!right) return left; // 右子树找不到，返回左子树
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return root; // 左右子树分别有一个，则返回root
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};
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```
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- Python Code:
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``` python
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# Definition for a binary tree node.
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# class TreeNode:
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# def __init__(self, x):
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# self.val = x
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# self.left = None
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# self.right = None
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class Solution:
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def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
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if not root or root == p or root == q:
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return root
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left = self.lowestCommonAncestor(root.left, p, q)
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right = self.lowestCommonAncestor(root.right, p, q)
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if not left:
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return right
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if not right:
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return left
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else:
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return root
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```
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## 扩展
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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？
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