Path Sum

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.

思路：后续遍历即可，只要访问到叶子节点，就把栈里的全部求和并与sum作比较，这里不能用stack，

因为stack不能访问栈中所有节点的值，所以用vector实现栈。

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if(NULL == root)
        return false;
        vector<TreeNode *> vint;
        TreeNode *last,*tmp;
        vint.push_back(root);
        tmp = root->left;
        last = NULL;
        int sumOfNodes = 0;
        vector<TreeNode*>::iterator iter;
        while(!vint.empty())
        {
            while(tmp != NULL)
            {
                vint.push_back(tmp);
                tmp = tmp->left;
            }
            tmp = vint.back();
            if(tmp->right == NULL)
            {
                if(tmp->left == NULL)
                {
                    sumOfNodes = 0;
                    for(iter = vint.begin();iter != vint.end();++iter)
                    {
                        sumOfNodes = sumOfNodes + (*iter)->val;
                        cout << sumOfNodes << endl;
                    }
                    if(sum == sumOfNodes)
                    return true;
                }
                last = vint.back();
                vint.pop_back();
            }
            else if(tmp->right != NULL && tmp->right == last)
            {
                last = vint.back();
                vint.pop_back();
            }
            if(!vint.empty() && vint.back()->right != NULL && vint.back()->right != last)
            tmp = vint.back()->right;
            else
            tmp = NULL;
        }
        return false;
        
    }
};