leetcode_question_102 Binary Tree Level Order Traversal

最新推荐文章于 2025-08-19 21:22:03 发布

原创最新推荐文章于 2025-08-19 21:22:03 发布 · 1.4k 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #Binary Tree Level Or #bfs

算法与数据结构同时被 3 个专栏收录

223 篇文章

订阅专栏

程序员笔试面试

223 篇文章

订阅专栏

Leetcode

179 篇文章

订阅专栏

本文介绍了一种使用广度优先搜索（BFS）策略来实现二叉树节点值的层次遍历算法。通过该算法可以按从左到右、逐层的顺序返回二叉树各节点的值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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,#,#,15,7},

return its level order traversal as:

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

BFS:

vector<vector<int> > levelOrder(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> matrix;
        if(root == NULL) return matrix;
		vector<int> r;
		r.push_back(root->val);
		matrix.push_back(r);

		vector<TreeNode *> path;
		path.push_back(root);
		
		int count = 1;
		while(!path.empty())
		{
			TreeNode* tn = path.front();
			if(tn->left) path.push_back(tn->left);
			if(tn->right) path.push_back(tn->right);
			path.erase(path.begin());
			count--;
			if(count==0)
			{
				vector<int> tmp;
				vector<TreeNode*>::iterator it = path.begin();
				for(;it != path.end(); ++it)
					tmp.push_back((*it)->val);
                if(tmp.size() > 0)
				    matrix.push_back(tmp);
				count = path.size();
			}
		}
		return matrix;
    }