112. 路径总和（非递归方法）

 bool hasPathSum(TreeNode* root, int sum) {
         if (root == NULL) 
            return false;
         if (root->val == sum && root->left ==  NULL && root->right == NULL) 
            return true;
         return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
    }

非递归的关键在于回退时在什么情况下减去节点的值，即该节点右子树已经遍历，要减去该节点值，回退到上一层 

bool hasPathSum(TreeNode *root, int sum) {
        stack s;
        TreeNode *pre = NULL, *cur = root;
        int SUM = 0;
        while (cur || !s.empty()) {
            while (cur) {
                s.push(cur);
                SUM += cur->val;
                cur = cur->left;
            }
            cur = s.top();
            if (cur->left == NULL && cur->right == NULL && SUM == sum) {
                return true;
            }
            if (cur->right && pre != cur->right) {
                cur = cur->right;
            } else {
                pre = cur;
                s.pop();
                SUM -= cur->val;
                cur = NULL;
            }
        }
        return false;
    }