LeetCode700. Search in a Binary Search Tree

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;
    }
};
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