[LeetCode] 129. Sum Root to Leaf Numbers

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.

本题将每条路径上的数字组合成一个大整数，然后求出这些整数的加和。

递归法：
左支（或右支）上得到的组合数字等于左子树（或右子树）的数字+根节点数字*10。
同理，将左节点或右节点看做根节点，求得该节点的组合数字。

代码如下：

class Solution {
public:
    int sumNumbers(TreeNode* root) {
        if (!root)return 0;
        if (!root->left && !root->right)return root->val;

        int res = 0;
        if(root->left)res += branchNum(root->left,root->val*10);
        if (root->right)res += branchNum(root->right, root->val * 10);
        return res;
    }
    int branchNum(TreeNode* root, int num) {
        if (!root->left && !root->right)return num + root->val;

        int res = 0;
        if (root->left)res += branchNum(root->left, (num + root->val) * 10);
        if (root->right)res += branchNum(root->right, (num + root->val) * 10);
        return res;
    }
};