145. Binary Tree Postorder Traversal

Given a binary tree, return the postorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

   1
    
     2
    /
   3

思路： 借助于一个栈，依次将根节点的右子节点和左子节点压入栈中。如果一个节点为叶子节点，或者前一个出栈的元素为当前栈顶节点的子节点，则出栈。

 vector<int> postorderTraversal(TreeNode* root) {
         vector<int> res;
         stack<TreeNode*> _stack;
         TreeNode *cur=root;
         TreeNode *pre=NULL;
         if(cur!=NULL)
             _stack.push(cur);
         while(!_stack.empty())
         {
             cur=_stack.top();
             if((cur->left==NULL&&cur->right==NULL)||(pre&&(cur->left==pre||cur->right==pre)))
             {
                 res.push_back(cur->val);
                 pre=cur;
                 _stack.pop();
             }
             else
             {
                 if(cur->right)
                     _stack.push(cur->right);
                 if(cur->left)
                     _stack.push(cur->left);
             }
         }
        return res;
    }