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
    /
   3
return [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;
    }
};