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?

递归解法：

class Solution {public:    void postorderRecursive(TreeNode *node, vector<int> &result)    {        if(node==NULL)        {            return ;        }                postorderRecursive(node->left, result);        postorderRecursive(node->right, result);                result.push_back(node->val);    }public:    vector<int> postorderTraversal(TreeNode *root) {        vector<int> result;        postorderRecursive(root, result);                return result;            }};

非递归解法:

/** * Definition for binary tree * 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> result;        stack<TreeNode *> stack_tree;        TreeNode *currentNode = root;                while(currentNode!=NULL)        {            stack_tree.push(currentNode);            currentNode = currentNode->left;        }        TreeNode *currentNode2;        while(!stack_tree.empty())        {            currentNode = stack_tree.top();            if(currentNode->right==NULL)            {                result.push_back(currentNode->val);                stack_tree.pop();            }            else            {                currentNode2 = currentNode->right;                currentNode->right = NULL;  //关键所在                //stack_tree.push(currentNode);                while(currentNode2!=NULL)                {                    stack_tree.push(currentNode2);                    currentNode2 = currentNode2->left;                }            }        }                return result;            }};