Path Sum

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.

题目解析：

（1）总体上讲只要层次遍历一次就可以了啊，当前出栈的节点为sum - 路径上的节点值的和。

因此当为0，且他没有左右子树的时候，那我们就得到了结果。

（2）同时不要忘了考虑边界条件，当树为空的情况。

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