129. Sum Root to Leaf Numbers

问题描述：

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：

可以用递归的方式来求解。

使用中序遍历。

判断每个节点是否为叶子节点。

若为叶子节点，则需要将目前的数字加入和中。

代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode* root) {
        if(!root) return 0;
        
        int sum = 0;
        traverse(root, 0, sum);
        return sum;
    }
    
    void traverse(TreeNode* root, int curVal, int &sum){
        int ret = curVal*10 + root->val;
        if(!root->right && !root->left){
            sum += ret;
            return;    
        }
        if(root->left) traverse(root->left, ret, sum);
        if(root->right) traverse(root->right, ret, sum);
        
    }
};