leetcode 102 Binary Tree Level Order Traversal-优快云博客

本文介绍了一种使用队列实现的二叉树层次遍历算法，并通过具体代码示例展示了如何构造二叉树并进行层次遍历，最后输出各层节点的值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include<iostream>
#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) {
		queue<TreeNode*> que;
		vector<vector<int>> res;
		vector<int> list;
		que.push(root);
		int last_num = 1;
		int cur_num = 0;
		while(!que.empty()){
			//vector<int> list;
			TreeNode *root = que.front();
			cout << "###### one #############" << endl;
			cout << "adding value is: " << root->val << endl;
			list.push_back(root->val);
			cout << "list size 1 is: " << list.size() << endl;
			que.pop();
			last_num--;
			if(root->left !=NULL){
				cout << "left val is: " << root->left->val << endl;
				que.push(root->left);		
				cur_num++;
			}
			if(root->right !=NULL){
				cout << "right val is: " << root->right->val << endl;
				que.push(root->right);		
				cur_num++;
			}
			if(last_num == 0){
				cout << "list size 2 is: " << list.size() << endl;
				res.push_back(list);
				last_num = cur_num;
				cur_num = 0;
				vector<int> list;
				cout << "list size 3 is: " << list.size() << endl;
			}
		}
		return res;
	}
};
int main(){
	solution so;
	TreeNode *t1 = new TreeNode(3);
	TreeNode *t2 = new TreeNode(9);
	TreeNode *t3 = new TreeNode(20);
	TreeNode *t4 = new TreeNode(15);
	TreeNode *t5 = new TreeNode(7);

	t1->left = t2;
	t1->right = t3;
	t3->left = t4;
	t3->right = t5;

	vector<vector<int>> res = so.levelOrder(t1);
	for(int i = 0;i < res.size();i++){
		for(int j=0; j<res[i].size();j++){
			cout << res[i][j] << ",";	
		}	
		cout <<endl; 
	}
}