[leetcode] Binary Tree Level Order Traversal

最新推荐文章于 2021-11-17 20:57:12 发布

wjk20120522

最新推荐文章于 2021-11-17 20:57:12 发布

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分类专栏： leetcode 文章标签： leetcode Binary Tree Level Order Traversal

本文链接：https://blog.youkuaiyun.com/jkwangbest/article/details/23363515

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本文介绍了一种使用两个队列实现二叉树层次遍历的方法，并通过具体示例展示了如何构建二叉树并进行层次遍历的过程。

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#include <iostream>
#include <ctime>
#include <string>
#include <cmath>
#include <vector>
#include <queue>
using namespace std;

/* the data structure of TreeNode */
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) {
		// use two STL queue to travel the tree 
		queue<TreeNode *> q,temp;
		vector<vector<int>> s;
		TreeNode * front_val;
		if(!root) {
			return s;
		}
		q.push(root);
		while(!q.empty()) {
			//store each level nodes in v
			vector<int> v;
			while(!q.empty()) {
				front_val = q.front();
				q.pop();
				if(front_val->left) {
					temp.push(front_val->left);
				}
				if(front_val->right) {
					temp.push(front_val->right);
				}
				v.push_back(front_val->val);
			}
			//put each v in s
			s.push_back(v);
			while(!temp.empty()) {
				front_val =  temp.front();
				temp.pop();
				q.push(front_val);
			}
		}
		return s;
	}
	
};

int main()
{
	Solution s;
	TreeNode * root = new TreeNode(3) ;
	TreeNode * two = new TreeNode(9) ;
	TreeNode * three = new TreeNode(20) ;
	TreeNode * four = new TreeNode(15) ;
	TreeNode * five = new TreeNode(7) ;

	root->left = two;
	root->right = three;
	three->left = four;
	three->right = five;
	vector<vector<int>> v;
	v = s.levelOrder(root);
	return 0;
}