leetcode - [6]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].

思路：后序遍历是按照“左子树，右子树，根”的顺序访问元素。那么根或者其它父亲元素就要先压入栈，然后再弹出。

#include <iostream>
#include <algorithm>
#include <vector>
#include <stack>

using namespace std;

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 *> s;
        if (!root) {
            return res;
        }
        s.push(root);
        while (!s.empty()) {
            TreeNode *p = s.top(); s.pop();
            res.push_back(p->val);

            if (p->right) {
                s.push(p->right);
            }

            if (p->left) {
                s.push(p->left);
            }
        }
        reverse(res.begin(), res.end());
        return res;
    }
};

int main(int argc, char *argv[]) {
    TreeNode *p = new TreeNode();
    p->right = new TreeNode();
    p->left = new TreeNode();    

    Solution *solution = new Solution();

    vector<int> res;
    res = solution->postorderTraversal(p);

    vector<int>::iterator it;
    for (it = res.begin(); it != res.end(); it++) {
        cout << *it << endl;
    }

}