700. Search in a Binary Search Tree*

700. Search in a Binary Search Tree*

https://leetcode.com/problems/search-in-a-binary-search-tree/

题目描述

Given the root node of a binary search tree (BST) and a value. You need to find the node in the BST that the node’s value equals the given value. Return the subtree rooted with that node. If such node doesn’t exist, you should return NULL.

For example,

Given the tree:
        4
       / \
      2   7
     / \
    1   3

And the value to search: 2
You should return this subtree:

      2     
     / \   
    1   3

In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.

Note that an empty tree is represented by NULL, therefore you would see the expected output (serialized tree format) as [], not null.

C++ 实现 1

很久没有写树有关的题目了, 先做一些简单的回忆一下基本概念. 二叉搜索树的定义就是左子树的所有节点的值小于根节点, 而右子树的所有节点的值大于根节点. 所以代码如下:

/**
 * 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* searchBST(TreeNode* root, int val) {
        if (!root) return nullptr;
        if (val < root->val) return searchBST(root->left, val);
        else if (val > root->val) return searchBST(root->right, val);
        else return root;
    }
};