## [Leetcode]235. Lowest Common Ancestor of a Binary Search Tree(C++)

### 题目描述

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

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 = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8` Output: `6` Explanation: The LCA of nodes 2 and 8 is 6.

#### 例子 2

Input: `root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4` Output: `2` Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

#### 例子 3

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

### 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

``````/**
* 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 == p || root == q) return root;
if (p->val < q->val) {
if (p->val < root->val && root->val < q->val)
return root;
else if (q->val < root->val) {
return lowestCommonAncestor(root->left, p, q);
} else {
return lowestCommonAncestor(root->right, p, q);
}
} else {
if (p->val > root->val && root->val > q->val) {
return root;
} else if (p->val < root->val) {
return lowestCommonAncestor(root->left, p, q);
} else {
return lowestCommonAncestor(root->right, p, q);
}
}
}
};
``````
• 时间复杂度: O(n)
• 空间复杂度: O(h)

GitHub 代码同步地址： 235.LowestCommonAncestorOfABinarySearchTree.cpp

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