129. Sum Root to Leaf Numbers #

Problem #

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.

Problem Summary #

Given a binary tree, each of its nodes stores a digit from 0-9, and each path from the root to a leaf node represents a number. For example, the path 1->2->3 from the root to a leaf node represents the number 123. Calculate the sum of all numbers generated from root-to-leaf paths. Note: A leaf node refers to a node with no child nodes.

Solution Approach #

• This problem is a variation of Problem 257. Problem 257 requires outputting each path from the root node to a leaf node. In this problem, each number from the root node to a leaf node is concatenated, and then each path is accumulated to find the final total sum. The actual solution idea is basically unchanged. Use the idea of preorder traversal: starting from the root node, keep adding until reaching a leaf node, and aggregate once at each leaf node.

Code #

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)
}