129. Sum Root to Leaf Numbers | 二叉树整数和

最新推荐文章于 2019-10-08 21:01:19 发布

原创最新推荐文章于 2019-10-08 21:01:19 发布 · 372 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

57 篇文章

订阅专栏

本文介绍了一种使用深度优先遍历的方法来解决二叉树中所有从根节点到叶子节点路径数字之和的问题。通过递归的方式遍历每个节点，并累加形成完整的路径数值，最终返回所有路径数值的总和。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Total Accepted: 105665
Total Submissions: 295189
Difficulty: Medium
Contributor: LeetCode

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   / \
  2   3

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

Subscribe to see which companies asked this question.

思路：深度优先遍历。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
  public int sumNumbers(TreeNode root) {
		int sumNumber = 0;
		int currNum = 0;
		return sumNumbersHelper(root, currNum);

	}

	public int sumNumbersHelper(TreeNode root, int currNum) {
		int sum  =0 ;
		if (root == null) {
			return sum;
		}
		if (root.left == null && root.right == null) {
			currNum = currNum * 10 + root.val;
			sum+=currNum;
			return sum;
		}
		int left =sumNumbersHelper(root.left, currNum * 10 + root.val);
		int right = sumNumbersHelper(root.right, currNum * 10 + root.val);
		return left+right;
	}
}