Leetcode 700 - Search in a binary search tree

Note:

  • Because it’s a binary search tree, we can decide whethere go left or right based on val. It’ll save lots of time instead of traversing all the nodes.
  • Use a flag to tell recursion that we’ve found a result so no more recursion.

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

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Input: root = [4,2,7,1,3], val = 2
Output: [2,1,3]
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/**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @param {number} val
* @return {TreeNode}
*/
var searchBST = function(root, val) {
let result = null;
let find = false;
traverse(root);
return result;

function traverse(node) {
if (!node || find) return;
if (node.val === val) {
result = node;
find = true;
return;
}
if (node.val > val) traverse(node.left);
if (node.val < val) traverse(node.right);
}
};