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.
解题思路1：
以按层遍历的思维，当要深入下一层时，将当前层累加的value值乘以10，并带入下一层继续运算，直到运算到叶子结点。之后累加所有叶子结点。
使用递归很容易实现：
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 int sumNumbers(TreeNode *root) { 13 if (!root) 14 return 0; 15 16 if (!root->left && !root->right) 17 return root->val; 18 else 19 return sumLeaf(root->left, root->val) + 20 sumLeaf(root->right, root->val); 21 } 22 23 int sumLeaf(TreeNode *node, int cur_sum) { 24 if (!node) 25 return 0; 26 27 if (!node->left & !node->right) 28 return cur_sum * 10 + node->val; 29 else 30 return sumLeaf(node->left, cur_sum * 10 + node->val) + 31 sumLeaf(node->right, cur_sum * 10 + node->val); 32 } 33 };
解题思路2：
递归的使用，造成程序运行慢。为了不使用递归可以按照二叉树先序遍历的方法。
代码：