Description#
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:
![](https://spcdn.pages.dev/leetcode/problems/0700.Search%20in%20a%20Binary%20Search%20Tree/images/tree1.jpg)
Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]
Example 2:
![](https://spcdn.pages.dev/leetcode/problems/0700.Search%20in%20a%20Binary%20Search%20Tree/images/tree2.jpg)
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
Solutions#
Solution 1#
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| # Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def searchBST(self, root: TreeNode, val: int) -> TreeNode:
if root is None or root.val == val:
return root
return (
self.searchBST(root.right, val)
if root.val < val
else self.searchBST(root.left, val)
)
|
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| /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public TreeNode searchBST(TreeNode root, int val) {
if (root == null || root.val == val) {
return root;
}
return root.val < val ? searchBST(root.right, val) : searchBST(root.left, val);
}
}
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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 || root->val == val) return root;
return root->val < val ? searchBST(root->right, val) : searchBST(root->left, val);
}
};
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| /**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func searchBST(root *TreeNode, val int) *TreeNode {
if root == nil || root.Val == val {
return root
}
if root.Val < val {
return searchBST(root.Right, val)
}
return searchBST(root.Left, val)
}
|