二叉树层次遍历

题目

如题

分析

	 第一层的下标其实就是0，level其实就是层数，level确认把每个数字放入对应的层里， 其实这种思想是一种BFS思想

解法1：递归

#include<stdio.h>
#include<iostream>
#include<stdlib.h>
#include<vector>
#include<queue>

using namespace std;

struct TreeNode{
	int val;
	TreeNode* left;
	TreeNode* right;
	TreeNode() :val(0), left(NULL), right(NULL){}
};

class MyClass
{
public:
	vector<vector<int>> levelOrder(TreeNode* root) {
		vector<vector<int>> res;
		recordLevelOrder(root, 0, res);
		return res;
	}

	void recordLevelOrder(TreeNode* root, int level, vector<vector<int>>& res){
		if (root == NULL){
			return;
		}

		if (res.size() == level){
			res.push_back({});
		}

		res[level].push_back(root->val);
		if (root->left){
			recordLevelOrder(root->left, level + 1, res);
		}
		if (root->right){
			recordLevelOrder(root->right, level + 1, res);
		}
		return;
	}
};

解法2：

class MyClass
{
public:
	vector<vector<int>> levelOrder(TreeNode* root) {
	
		vector<vector<int>> res;
		if (root == NULL)
		{
			return res;
		}
		queue<TreeNode*> st;
	

		st.push(root);
		while (!st.empty())
		{
			vector<int> tmp;
			int size = st.size();
			for (int i = 0; i < size; i++)
			{

				TreeNode* node = st.front();
				st.pop();
				int val = node->val;
				tmp.push_back(val);
				if (node->left)
				{
					st.push(node->left);
				}
				if (node->right)
				{
					st.push(node->right);
				}
			}
			res.push_back(tmp);
		}
		return res;
	}
};