0129. Sum Root to Leaf Numbers

129. Sum Root to Leaf Numbers #

题目 #

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.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

题目大意 #

给定一个二叉树,它的每个结点都存放一个 0-9 的数字,每条从根到叶子节点的路径都代表一个数字。例如,从根到叶子节点路径 1->2->3 代表数字 123。计算从根到叶子节点生成的所有数字之和。说明: 叶子节点是指没有子节点的节点。

解题思路 #

  • 这一题是第 257 题的变形题,第 257 题要求输出每条从根节点到叶子节点的路径,这一题变成了把每一个从根节点到叶子节点的数字都串联起来,再累加每条路径,求出最后的总和。实际做题思路基本没变。运用前序遍历的思想,当从根节点出发一直加到叶子节点,每个叶子节点汇总一次。

代码 #


package leetcode

import (
	"github.com/halfrost/LeetCode-Go/structures"
)

// TreeNode define
type TreeNode = structures.TreeNode

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

func sumNumbers(root *TreeNode) int {
	res := 0
	dfs(root,0,&res)
	return res
}

func dfs(root *TreeNode,sum int,res *int)  {
	if root == nil{
		return
	}
	sum = sum*10 + root.Val
	if root.Left == nil && root.Right == nil{
		*res += sum
		return
	}
	dfs(root.Left,sum,res)
	dfs(root.Right,sum,res)
}


⬅️上一页

下一页➡️

Calendar Nov 26, 2021
Edit Edit this page
本站总访问量:  次 您是本站第  位访问者