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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},
{3,9,20,#,#,15,7}
3 / \ 9 20 / \ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
层序遍历二叉树是典型的广度优先搜索 BFS 的应用,但是这里稍微复杂一点的是,要把各个层的数分开,存到一个二维向量里面,大体思路还是基本相同的,建立一个 queue,然后先把根节点放进去,这时候找根节点的左右两个子节点,这时候去掉根节点,此时 queue 里的元素就是下一层的所有节点,用一个 for 循环遍历它们,然后存到一个一维向量里,遍历完之后再把这个一维向量存到二维向量里,以此类推,可以完成层序遍历,参见代码如下:
解法一:
class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { if (!root) return {}; vector<vector<int>> res; queue<TreeNode*> q{{root}}; while (!q.empty()) { vector<int> oneLevel; for (int i = q.size(); i > 0; --i) { TreeNode *t = q.front(); q.pop(); oneLevel.push_back(t->val); if (t->left) q.push(t->left); if (t->right) q.push(t->right); } res.push_back(oneLevel); } return res; } };
下面来看递归的写法,核心就在于需要一个二维数组,和一个变量 level,关于 level 的作用可以参见博主的另一篇博客 Binary Tree Level Order Traversal II 中的讲解,参见代码如下:
解法二:
class Solution { public: vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int>> res; levelorder(root, 0, res); return res; } void levelorder(TreeNode* node, int level, vector<vector<int>>& res) { if (!node) return; if (res.size() == level) res.push_back({}); res[level].push_back(node->val); if (node->left) levelorder(node->left, level + 1, res); if (node->right) levelorder(node->right, level + 1, res); } };
Github 同步地址:
#102
类似题目:
Binary Tree Level Order Traversal II
Binary Tree Zigzag Level Order Traversal
Minimum Depth of Binary Tree
Binary Tree Vertical Order Traversal
Average of Levels in Binary Tree
N-ary Tree Level Order Traversal
参考资料:
https://leetcode.com/problems/binary-tree-level-order-traversal/
https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/33445/Java-Solution-using-DFS
https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/33450/Java-solution-with-a-queue-used
https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/114449/A-general-approach-to-level-order-traversal-questions-in-Java
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered:
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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:
层序遍历二叉树是典型的广度优先搜索 BFS 的应用,但是这里稍微复杂一点的是,要把各个层的数分开,存到一个二维向量里面,大体思路还是基本相同的,建立一个 queue,然后先把根节点放进去,这时候找根节点的左右两个子节点,这时候去掉根节点,此时 queue 里的元素就是下一层的所有节点,用一个 for 循环遍历它们,然后存到一个一维向量里,遍历完之后再把这个一维向量存到二维向量里,以此类推,可以完成层序遍历,参见代码如下:
解法一:
下面来看递归的写法,核心就在于需要一个二维数组,和一个变量 level,关于 level 的作用可以参见博主的另一篇博客 Binary Tree Level Order Traversal II 中的讲解,参见代码如下:
解法二:
Github 同步地址:
#102
类似题目:
Binary Tree Level Order Traversal II
Binary Tree Zigzag Level Order Traversal
Minimum Depth of Binary Tree
Binary Tree Vertical Order Traversal
Average of Levels in Binary Tree
N-ary Tree Level Order Traversal
参考资料:
https://leetcode.com/problems/binary-tree-level-order-traversal/
https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/33445/Java-Solution-using-DFS
https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/33450/Java-solution-with-a-queue-used
https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/114449/A-general-approach-to-level-order-traversal-questions-in-Java
LeetCode All in One 题目讲解汇总(持续更新中...)
The text was updated successfully, but these errors were encountered: