Path Sum

Q：

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.

A:

额，仍然是后序，好吧，有一点小变化，看来写一次代码能套用n多次。

/**
 * Definition for binary tree
 * 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) {
    // Start typing your C/C++ solution below
    // DO NOT write int main() function
    if (!root) return false; 
    stack<TreeNode*> s;
    stack<bool> r;
    int cur_sum = 0;
    TreeNode* cur = root;
    while (cur || !s.empty()) {
      while (cur) {
        s.push(cur);
        r.push(false);
        cur_sum += cur->val;
        cur = cur->left;
      }   
      while (!r.empty() && r.top()) {
        TreeNode* tmp = s.top();
        if (!tmp->left && !tmp->right) {
          if (cur_sum == sum) return true;
        }   
        cur_sum -= tmp->val;
        s.pop();
        r.pop();
      }   
          
      if (!s.empty()) {
        cur = s.top()->right;
        r.top() = true;
      }   
    }   
    return false;
  }
};