LeetCode｜700. Search in Binary Search Tree

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 <= 10^7
• root is a binary search tree.
• 1 <= val <= 10^7

思路

简单不考虑BST的特性，直接遍历全部二叉树的所有节点。
当找到目标节点后，即直接返回。

代码

/**
 * 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:
    void searchNode(TreeNode* root, int val, TreeNode*& node) {
        if (node != NULL)
            return;
        if (root->val == val)
        {
            node = root;
            return;
        }
        if (root->left != NULL) {
            searchNode(root->left, val, node);
        } 
        if (root->right != NULL) {
            searchNode(root->right, val, node);
        } 
        return;
    }

    TreeNode* searchBST(TreeNode* root, int val) {
        if (root == NULL)
        {
            return root;
        }

        TreeNode* node = NULL;
        searchNode(root, val, node);
        return node;
    }
};