(Easy) Search in Binary Tree - LeetCode

2019-08-29

(Easy) Search in Binary Tree - LeetCode

Description: 

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.

Accepted
75,536
Submissions
109,282

 

Solution:

using Reccrusion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
       if(root == null || val == root.val)
            return root;
        else
            return val < root.val ? searchBST(root.left, val) : searchBST(root.right, val);

     
    }
}

 

(Easy) Search in Binary Tree - LeetCode

(Easy) Search in Binary Tree - LeetCode

原文地址:https://www.cnblogs.com/codingyangmao/p/11429672.html