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Lintcode482-Binary Tree Level Sum-Easy

时间:2019-04-06 14:13:23      阅读:117      评论:0      收藏:0      [点我收藏+]

标签:bsp   int   val   NPU   lin   eve   treenode   solution   markdown   

482. Binary Tree Level Sum

Given a binary tree and an integer which is the depth of the target level.

Calculate the sum of the nodes in the target level.

Example

Example 1:

Input:

     1
   /     2     3
 / \   / 4   5 6   7
   /         8         9
and depth=2
Output:5 

Example 2:

Input:

     1
   /     2     3
 / \   / 4   5 6   7
   /         8         9
and depth=3
Output:22

 
思路:递归法

注意:

  1. sum 定义为类成员变量

代码:

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: the root of the binary tree
     * @param level: the depth of the target level
     * @return: An integer
     */
    int sum = 0;
    public int levelSum(TreeNode root, int level) {
       helper(root, 1, level);
       return sum;
    }
    public void helper(TreeNode root, int depth, int level) {
        if (root == null) {
            return;
        }
        
        if (depth == level) {
            sum += root.val;
            return;
        }
        
        helper(root.left, depth + 1, level);
        helper(root.right, depth + 1, level);
    }
}

 

Lintcode482-Binary Tree Level Sum-Easy

标签:bsp   int   val   NPU   lin   eve   treenode   solution   markdown   

原文地址:https://www.cnblogs.com/Jessiezyr/p/10661423.html

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