Problem
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
Solution (Java)
/**
* 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) {
while (root != null && root.val != val) {
if (root.val > val) {
root = root.left;
} else {
root = root.right;
}
}
return root;
}
}
Explain:
nope.
Complexity:
- Time complexity : O(n).
- Space complexity : O(n).