[LeetCode] Sum Root to Leaf Numbers, Solution_maximum sum root to leaf path-优快云博客

本文介绍了一种通过递归算法解决二叉树所有根到叶子节点路径数值之和的方法。利用递归函数遍历二叉树，每条路径转换为数值并累加求和，最终返回所有路径数值的总和。

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.

» Solve this problem

[Thoughts]
Recursion. Similar as [LeetCode] Binary Tree Maximum Path Sum Solution, the difference here is only adding a track variable to sum all the paths.

[Code]

1:       int sumNumbers(TreeNode *root) {  
2:            int sum=0, path =0;            
3:            GenerateSum(root, sum, path);  
4:            return sum;  
5:       }  
6:       void GenerateSum(TreeNode *root, int& sum, int path)  
7:       {  
8:            if(root == NULL) return;      
9:            path = path*10 +root->val;  
10:            if(root->left == NULL && root->right == NULL)  
11:            {  
12:                 sum+=path;  
13:                 return;  
14:            }  
15:            GenerateSum(root->left, sum, path);  
16:            GenerateSum(root->right, sum, path);  
17:       }