## [Leetcode]236. Lowest Common Ancestor of a Binary Tree(C++)

### 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

``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/

class Solution {
public:
TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {
if (root == nullptr) return nullptr;
if (root == p || root == q) return root;
// search node in left subtree
TreeNode* left_search = lowestCommonAncestor(root->left, p, q);
TreeNode* right_search = lowestCommonAncestor(root->right, p, q);
// we find node in both subtree
if (left_search && right_search) {
return root;
} else if (left_search) {
return left_search;
} else {
return right_search;
}
}
};
``````
• 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

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