//

// Solution102.hpp

// Algorithm

//

// Created by Pancf on 2019/12/23.

// Copyright © 2019 Pancf. All rights reserved.

//

#ifndef Solution102_hpp

#define Solution102_hpp

#include <stdio.h>

#include <vector>

using std::vector;

class Solution102 {

/**

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:

Given binary tree [3,9,20,null,null,15,7],

3

/ \

9 20

/ \

15 7

return its level order traversal as:

[

[3],

[9,20],

[15,7]

]

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

Accept details:

Runtime: 8 ms, faster than 58.20% of C++ online submissions for Binary Tree Level Order Traversal.

Memory Usage: 13.9 MB, less than 77.46% of C++ online submissions for Binary Tree Level Order Traversal.

思路：二叉树的按层遍历，借用一个队列就完事了

*/

public:

struct TreeNode {

int val;

TreeNode *left;

TreeNode *right;

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

};

vector<vector<int>> levelOrder(TreeNode* root);

static void test() {

TreeNode root{3};

TreeNode node1{9};

TreeNode node2{20};

TreeNode node3{15};

TreeNode node4{7};

root.left = &node1;

root.right = &node2;

node2.left = &node3;

node2.right = &node4;

Solution102 s = Solution102();

auto rv = s.levelOrder(&root);

for (auto level_vec : rv) {

for (auto i : level_vec) {

printf("%d ", i);

}

printf("\n");

}

}

};

#endif /* Solution102_hpp */