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
return its level order traversal as:
[ [3], [9,20], [15,7] ]
層序遍歷二叉樹是典型的廣度優先搜索 BFS 的應用,但是這里稍微復雜一點的是,要把各個層的數分開,存到一個二維向量里面,大體思路還是基本相同的,建立一個 queue,然后先把根節點放進去,這時候找根節點的左右兩個子節點,這時候去掉根節點,此時 queue 里的元素就是下一層的所有節點,用一個 for 循環遍歷它們,然后存到一個一維向量里,遍歷完之后再把這個一維向量存到二維向量里,以此類推,可以完成層序遍歷,參見代碼如下:
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; } };
下面來看遞歸的寫法,核心就在於需要一個二維數組,和一個變量 level,關於 level 的作用可以參見博主的另一篇博客 Binary Tree Level Order Traversal II 中的講解,參見代碼如下:
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:
https://github.com/grandyang/leetcode/issues/102
類似題目:
Binary Tree Level Order Traversal II
Binary Tree Zigzag Level Order Traversal
Binary Tree Vertical Order Traversal
Average of Levels in Binary Tree
N-ary Tree Level Order Traversal
參考資料:
https://leetcode.com/problems/binary-tree-level-order-traversal/