//

// Solution145.hpp

// Algorithm

//

// Created by Pancf on 2019/12/21.

// Copyright © 2019 Pancf. All rights reserved.

//

#ifndef Solution145_hpp

#define Solution145_hpp

#include <stdio.h>

#include <vector>

using std::vector;

class Solution145 {

/**

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

Example:

Input: [1,null,2,3]

1

\

2

/

3

Output: [3,2,1]

Follow up: Recursive solution is trivial, could you do it iteratively?

================================================================================================

Accept details:

Runtime: 4 ms, faster than 59.15% of C++ online submissions for Binary Tree Postorder Traversal.

Memory Usage: 9.1 MB, less than 93.55% of C++ online submissions for Binary Tree Postorder Traversal.

思路：迭代法的后续遍历要比前序、中序稍难一点，具体思路见实现文件的注释

*/

public:

struct TreeNode {

int val;

TreeNode *left;

TreeNode *right;

TreeNode(int x) : val(x), left(NULL), right(NULL) {}

};

vector<int> postorderTraversal(TreeNode* root);

static void test() {

TreeNode root{1};

TreeNode node1{2};

TreeNode node2{3};

root.right = &node1;

node1.left = &node2;

Solution145 s = Solution145();

auto rv = s.postorderTraversal(&root);

for (auto ele : rv) {

printf("%d ", ele);

}

}

};

#endif /* Solution145_hpp */