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：
/**
* Definition for binary tree
* 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 ; if (!root->left && !root->right)
return root->val;
else
return sumLeaf(root->left, root->val) +
sumLeaf(root->right, root->val);
} int sumLeaf(TreeNode *node, int cur_sum) {
if (!node)
return ; if (!node->left & !node->right)
return cur_sum * + node->val;
else
return sumLeaf(node->left, cur_sum * + node->val) +
sumLeaf(node->right, cur_sum * + node->val);
}
};
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：