145. Binary Tree Postorder Traversal

/*Given a binary tree, return the postorder traversal of its nodes' values.For example:Given binary tree {1,#,2,3},   1    \     2    /   3return [3,2,1].Note: Recursive solution is trivial, could you do it iteratively?使用迭代实现二叉树的后序遍历*//** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    vector<int> postorderTraversal(TreeNode* root) {        vector<int> res;        stack<TreeNode*> st;        TreeNode* lastVisit=NULL;        TreeNode* ptr=root;        while(ptr || !st.empty())        {            if(ptr)            {                st.push(ptr);                ptr=ptr->left;            }            else            {                TreeNode* topNode=st.top();                if(topNode->right && topNode->right!=lastVisit)                    ptr=topNode->right;                else                {                    res.push_back(topNode->val);                    lastVisit=topNode;                    st.pop();                }            }        }        return res;    }    vector<int> postorderTraversal2(TreeNode* root) {        vector<int> res;        stack<TreeNode*> st;        if(root) st.push(root);        while(!st.empty())        {            TreeNode* topNode=st.top();            st.pop();            res.push_back(topNode->val);            if(topNode->left) st.push(topNode->left);            if(topNode->right) st.push(topNode->right);        }        reverse(res.begin(),res.end());        return res;    }};