LeetCode--Sum Root to Leaf Numbers

题目：

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. 

思路：类似于二叉树的遍历，遍历的过过程中要记录节点数据，并且转化为sum（越接近叶子节点的节点的数值离各位最近，因为叶子节点是个位数，叶子节点的根节点是十位数。。。根root节点更高位），要标记已经计算过的节点（将值赋值为-1），叶子节点遍历过回溯到根节点，继续搜索。

#include<iostream>
#include<stack>
using namespace std;
//二叉树的定义 
struct TreeNode {int val;TreeNode *left;TreeNode *right;TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
int sumNumbers(TreeNode* root)
{TreeNode* p = root;if(p==NULL) return 0; //栈，搜索节点时，存储节点 stack<TreeNode*> s;//根到每一个叶子节点的和 int sum = 0;//sum 之和 int result = 0;//最左下的节点（不一定是叶子） while(p!=NULL){s.push(p);sum = sum * 10 + p->val;p = p->left;}while(!s.empty()){p = s.top();//不是叶子节点，且没遍历过 if(p->right!=NULL&&p->right->val!=-1){p = p->right;s.push(p);sum = sum*10 + p->val;while(p->left!=NULL){p = p->left;s.push(p);sum = sum*10 + p->val;}}else if(p->left==NULL && p->right == NULL)//叶子节点 {result += sum;//回溯到该节点的根 s.pop();sum/=10;//标记 p->val = -1;}//不是叶子节点，且已遍历过 else if((p->left==NULL&&p->right!=NULL&&p->right->val==-1)||(p->right==NULL&&p->left!=NULL&&p->left->val==-1)||(p->left!=NULL&&p->right!=NULL&&p->left->val==-1&&p->right->val==-1)){s.pop();sum/=10;//标记p->val = -1; }}return result;
}
int main(void)
{TreeNode* root = new TreeNode(9);TreeNode* p2 = new TreeNode(2);TreeNode* p3 = new TreeNode(3);TreeNode* p4 = new TreeNode(4);TreeNode* p5 = new TreeNode(5);TreeNode* p6 = new TreeNode(6);TreeNode* p7 = new TreeNode(7);root->left = p2;root->right = p3;p2->left = p4;p2->right = p5;p3->left = p6;p3->right = p7;int a = sumNumbers(root);cout<<a<<endl;delete(root);delete(p2);delete(p3);delete(p4);delete(p5);delete(p6);delete(p7);system("pause");return 0;
}

可以找一个比较“变态”的树，走一遍程序，思路就会很清楚。