0700 - Search in a Binary Search Tree

0700 - Search in a Binary Search Tree

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.

Examples

Input: root = [4,2,7,1,3], val = 2 Output: [2,1,3] 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

Java Solution

Recursive

/**
 * 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 val< root.val ? searchBST(root.left, val) : searchBST(root.right, val);
    }
}

Iterative

class Solution {
  public TreeNode searchBST(TreeNode root, int val) {
    while (root != null && val != root.val)
      root = val < root.val ? root.left : root.right;
    return root;
  }
}