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;}
};