package com.fishercoder.solutions;

import com.fishercoder.common.classes.TreeNode;

/**
 * 700. Search in a Binary Search Tree
 *
 * Given the root node of a binary search tree (BST) and a value.
 * You need to find the node in the BST that the node's value equals the given value.
 * Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.
 *
 * For example,
 *
 * Given the tree:
 *         4
 *        / \
 *       2   7
 *      / \
 *     1   3
 *
 * And the value to search: 2
 * You should return this subtree:
 *
 *       2
 *      / \
 *     1   3
 * In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.
 *
 * Note that an empty tree is represented by NULL,
 * therefore you would see the expected output (serialized tree format) as [], not null.
 * */
public class _700 {
    public static class Solution1 {
        public TreeNode searchBST(TreeNode root, int val) {
            if (root == null) {
                return null;
            } else if (root.val == val) {
                return root;
            } else if (root.val > val) {
                return searchBST(root.left, val);
            } else {
                return searchBST(root.right, val);
            }
        }
    }
}

@@ -82,786 +82,8 @@
 e;%0A%0A
-/**%0A * 700. Search in a Binary Search Tree%0A *%0A * Given the root node of a binary search tree (BST) and a value.%0A * You need to find the node in the BST that the node's value equals the given value.%0A * Return the subtree rooted with that node. If such node doesn't exist, you should return NULL.%0A *%0A * For example,%0A *%0A * Given the tree:%0A *         4%0A *        / %5C%0A *       2   7%0A *      / %5C%0A *     1   3%0A *%0A * And the value to search: 2%0A * You should return this subtree:%0A *%0A *       2%0A *      / %5C%0A *     1   3%0A * In the example above, if we want to search the value 5, since there is no node with value 5, we should return NULL.%0A *%0A * Note that an empty tree is represented by NULL,%0A * therefore you would see the expected output (serialized tree format) as %5B%5D, not null.%0A * */%0A
 publ

refactor 700