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### 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).”

### 例子

#### 例子 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.

#### 例子 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.

#### 例子 3

Input: `root = [1,2], p = 1, q = 2`
Output: 1

### Constraints

• The number of nodes in the tree is in the range `[2, 10^5]`.
• `-10^9 <= Node.val <= 10^9`
• All `Node.val` are unique.
• `p != q`
• `p` and `q` will exist in the BST.

### 解题思路

• 对一棵树同时搜索 `p` 或者 `q`
• 如果根结点是空，直接返回空指针
• 如果根结点是 `p` 或者 `q`，直接返回根结点（因为假如另一个节点在同一棵树内的话（题目保证了两个节点都会存在），最深的祖先节点一定是根结点）
• 否则对左右两颗子树进行同样操作，得到返回结果 `left``right`
• 假如 `left``right` 都非空，表示在左子树和右子树都搜到了其中一个节点，此时根结点为最近祖先节点，返回根结点
• 假如 `left``right` 只有一个非空节点，表示除了该非空节点的子树外，其他节点都不包含另一个节点，表明另一个节点一定在该节点的子树下，返回该节点作为最近祖先节点
• 假如 `left``right` 都为节点：不可能，因为题目保证两个节点都存在

• 时间复杂度: O(n)
• 空间复杂度: O(h) — 递归深度

GitHub 代码同步地址： 236.LowestCommonAncestorOfABinaryTree.cpp

GitHub: Leetcode-C++-Solution