700. Search in a Binary Search Tree

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700. Search in a Binary Search Tree

Description

Difficulty: Easy

Related Topics: Tree, Binary Search Tree, Binary Tree

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:

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Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]

Example 2:

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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

Hints/Notes

  • Use the property of BST

Solution

Language: C++

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/**
 * 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 == nullptr) return root;
        if (root->val < val) {
            return searchBST(root->right, val);
        }
        if (root->val > val) {
            return searchBST(root->left, val);
        }
        return root;
    }
};