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;}
};