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Solution102.hpp
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71 lines (63 loc) · 1.68 KB
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//
// 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 */