Path Sum

【题目描述】

Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.

For example:
Given the below binary tree and sum = 22,

              5             / \            4   8           /   / \          11  13  4         /  \      \        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

【思路】

分别实现了非递归和递归。

递归比较容易，用dfs就可以了。

非递归：使用两个堆栈，一个记录节点，一个记录到当前节点的路径节点总和与sum的差，当sum为0且该节点的左右子树均为空时为true。

【代码】

递归：

class Solution {public:    bool hasPathSum(TreeNode* root, int sum) {        return pathsum(root,sum,0); }bool pathsum(TreeNode* node,int sum,int nowsum){     if(node==NULL) return false;     if(node->left==NULL&&node->right==NULL) {         if(nowsum+node->val==sum) return true;         else return false;     }     return pathsum(node->left,sum,nowsum+node->val)||pathsum(node->right,sum,nowsum+node->val); }};

非递归：

/** * 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:    bool hasPathSum(TreeNode* root, int sum) {        stack<TreeNode*> s;        stack<int> snum;        if(root==NULL) return false;        s.push(root);        snum.push(sum);        while(!s.empty()){            TreeNode* node=s.top();            sum=snum.top()-node->val;            s.pop();            snum.pop();            if(sum==0&&node->left==NULL&&node->right==NULL) return true;             if(node->left){                if(node->right) {s.push(node->right);snum.push(sum);}                 s.push(node->left); snum.push(sum);            }             else if(node->right){                s.push(node->right);                snum.push(sum);            }        }        return false;    }};