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：典型的二叉树数据结构方法。非递归方式，可用两个栈，S和total，S记录节点，total记录从根节点到当前节点的和，当到达叶子节点时，判断其和是否等于sum。
class Solution {
public:
bool hasPathSum(TreeNode *root, int sum) {
stack<TreeNode*> S;
stack<int> total;
int val = 0;
while(root != NULL || !S.empty())
{
while(root != NULL)
{
S.push(root);
total.push(val + root->val);
val = total.top();
if (root->left == NULL && root->right == NULL) //是否到达叶子节点
{
if (total.top() == sum)
{
return true;
}
}
root = root->left;
}
root = S.top();
S.pop();
val = total.top();
total.pop();
root = root->right;
}
return false;
}
};