sum-root-to-leaf-numbers

1、题目描述：

Given a binary tree containing digits from0-9only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path1->2->3which represents the number123.
Find the total sum of all root-to-leaf numbers.
For example,
    1
   / \
  2   3


The root-to-leaf path1->2represents the number12.
The root-to-leaf path1->3represents the number13.
Return the sum = 12 + 13 =25.

2、代码实现：

1）、使用递归的方式：

/**
 * 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) {
		return cal_sum(root, 0);
	}

	int cal_sum(TreeNode *root, int sum)
	{
		//只有左右子树都为空时，才产生一个数字；像左子树为空，但右子树不为空的情况，则数字朝右边继续生长，
		//左子树不单独形成一个数字，所以root为NULL是返回0而不是sum，只有左右子树都为空时（叶子节点）一个数字才生成并返回
		if (!root)  
			return 0;
		if (!root->left && !root->right)  
			return sum * 10 + root->val;


		return cal_sum(root->left, sum * 10 + root->val) + cal_sum(root->right, sum * 10 + root->val);
	}
};

2）、也可以使用引用的方式：

/**
 * 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==NULL)
            return 0;
        int cur=0,sum=0;
        CalSum(root,cur,sum);
        return sum;
    }
     
    void CalSum(TreeNode* root,int cur,int& sum){
        cur=cur*10+root->val;
        if(root->left==NULL && root->right==NULL){
            sum+=cur;
            return;
        }
        if(root->left!=NULL)
            CalSum(root->left,cur,sum);
        if(root->right!=NULL)
            CalSum(root->right,cur,sum);
    }
};