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.
``````

## 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)
}

``````

Apr 8, 2023