leetcode 102. Binary Tree Level Order Traversal BFS广度优先遍历

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]
]

题意很简单，就是一个BFS广度优先遍历的求解，注意把握好每一层的size。

代码如下：

import java.util.ArrayList;import java.util.LinkedList;import java.util.List;import java.util.Queue;/*class TreeNode {     int val;     TreeNode left;     TreeNode right;     TreeNode(int x) { val = x; }}*/public class Solution {    public List<List<Integer>> levelOrder(TreeNode root)    {        List<List<Integer>> res=new ArrayList<>();        if(root==null)            return res;        Queue<TreeNode> queue=new LinkedList<>();        queue.add(root);        while(queue.isEmpty()==false)        {            int size=queue.size();            List<Integer> one=new ArrayList<>();            for(int i=0;i<size;i++)            {                TreeNode t=queue.poll();                one.add(t.val);                if(t.left!=null)                    queue.add(t.left);                if(t.right!=null)                    queue.add(t.right);            }            res.add(one);        }        return res;    }}

下面是C++的做法，就是一个简单的BFS广度优先遍历的做法

代码如下：

#include<iostream>#include <vector>#include <queue>using namespace std;/*struct TreeNode{    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x) : val(x), left(NULL), right(NULL) {}};*/class Solution {public:    vector<vector<int>> levelOrder(TreeNode* root)     {        vector<vector<int>> res;        if (root == NULL)            return res;        queue<TreeNode*> que;        que.push(root);        while (que.empty() == false)        {            int size = que.size();            vector<int> one;            for (int i = 0; i < size; i++)            {                TreeNode* top = que.front();                que.pop();                one.push_back(top->val);                if(top->left!=NULL)                    que.push(top->left);                if (top->right != NULL)                    que.push(top->right);            }            res.push_back(one);        }        return res;    }};