Binary Tree Level Order Traversal II (leetcode)

二叉树层次遍历

最新推荐文章于 2019-08-19 21:26:05 发布

原创最新推荐文章于 2019-08-19 21:26:05 发布 · 406 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#树遍历 #树层次遍历 #leetcode

leetcode题解专栏收录该内容

70 篇文章

订阅专栏

本文介绍了一种解决LeetCode上二叉树层次遍历II问题的方法，通过使用队列进行层次遍历，并在最后反转结果来实现从叶节点到根节点的层次遍历。

题目：

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree {3,9,20,#,#,15,7},

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

题目来源：https://oj.leetcode.com/problems/binary-tree-level-order-traversal-ii/

解题思路：层次遍历，然后调用一次reverse即可

#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;

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

vector<vector<int> > levelOrderBottom(TreeNode *root)
{
	vector<vector<int> >results;
	if(root==NULL)
		return results;
	queue<TreeNode *> q;
	q.push(root);
	int k=0,level=1;
	while(!q.empty())
	{
		k=level;
		level=0;
		vector<int> result;
		for(int i=0;i<k;i++)
		{
			result.push_back(q.front()->val);
			if(q.front()->left!=NULL)
			{
				q.push(q.front()->left);
				level++;
			}
			if(q.front()->right!=NULL)
			{
				q.push(q.front()->right);
				level++;
			}
			q.pop();
		}
		results.push_back(result);
	}
	reverse(results.begin(),results.end());
	return results;
}

int main()
{
	TreeNode* root=new TreeNode(1);
	root->left=new TreeNode(2);
	root->right=new TreeNode(3);
	root->left->left=new TreeNode(4);
	root->right->left=new TreeNode(5);
	vector<vector<int> > results=levelOrderBottom(root);

	system("pause");
	return 0;
}