文章目录
- 一、题目
- 二、题解
一、题目
You are given the root of a binary search tree (BST) and an integer val.
Find the node in the BST that the node’s value equals val and return the subtree rooted with that node. If such a node does not exist, return null.
Example 1:
Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]
Example 2:
Input: root = [4,2,7,1,3], val = 5
Output: []
Constraints:
The number of nodes in the tree is in the range [1, 5000].
1 <= Node.val <= 107
root is a binary search tree.
1 <= val <= 107
二、题解
/*** Definition for a binary tree node.* struct TreeNode {* int val;* TreeNode *left;* TreeNode *right;* TreeNode() : val(0), left(nullptr), right(nullptr) {}* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:TreeNode* searchBST(TreeNode* root, int val) {if(!root || root->val == val) return root;TreeNode* res = nullptr;if(root->val > val) res = searchBST(root->left,val);if(root->val < val) res = searchBST(root->right,val);return res;}
};