【leetcode刷题】T125-求根到叶子节点数字之和

示例 1:
输入: [1,2,3]
    1
   / \
  2   3
输出: 25
解释:
从根到叶子节点路径 1->2 代表数字 12.
从根到叶子节点路径 1->3 代表数字 13.
因此，数字总和 = 12 + 13 = 25.

示例 2:
输入: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
输出: 1026
解释:
从根到叶子节点路径 4->9->5 代表数字 495.
从根到叶子节点路径 4->9->1 代表数字 491.
从根到叶子节点路径 4->0 代表数字 40.
因此，数字总和 = 495 + 491 + 40 = 1026.

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def sumNumbers(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        sum0 = 0
        sum0 = self.get_value(root, '', sum0)
        return sum0

    def get_value(self, node, cur, sum0):
        if not node:
            return 0
        if not node.left and not node.right:
            cur += str(node.val)
            sum0 += int(cur)
            return sum0
        cur += str(node.val)
        return self.get_value(node.left, cur, sum0) + self.get_value(node.right, cur, sum0)

/**
 * 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) {
        int sum = 0;
        string cur = "";
        get_val(root, cur, sum);
        return sum;
    }

    void get_val(TreeNode* node, string cur, int& sum){
        if(!node)
            return;
        if(!node->left && !node->right){
            cur += (node->val + '0');
            sum += stoi(cur);
            return;
        }
        cur += (node->val + '0');
        get_val(node->left, cur, sum);
        get_val(node->right, cur, sum);
    }
};