【LeetCode】102. 二叉树的层序遍历

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【LeetCode】102. 二叉树的层序遍历

题目链接

文章目录

• • Python3
  • • 方法一： 广度优先搜索 (BFS) ⟮ O ( n ) ⟯ \lgroup O(n) \rgroup ⟮O(n)⟯
    • 方法二： 深度优先搜索 (DFS) ⟮ O ( n ) ⟯ \lgroup O(n) \rgroup ⟮O(n)⟯
  • C++
  • • 方法一： 广度优先搜索 (BFS) ⟮ O ( n ) ⟯ \lgroup O(n) \rgroup ⟮O(n)⟯

Python3

方法一： 广度优先搜索 (BFS) ⟮ O ( n ) ⟯ \lgroup O(n) \rgroup ⟮O(n)⟯

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:if not root:return []res = []queue =   [root,] while queue: tmp = [node.val for node in queue]res.append(tmp) # 取当前层 的结点值lis = [] ## 下一层 的结点for node in queue:if node.left:lis.append(node.left)if node.right:lis.append(node.right)queue = lis return res

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:if not root:return []ans = []dq = deque([root])while dq: level = [] ## 当前 遍历层n = len(dq)for _ in range(n):cur = dq.popleft()level.append(cur.val)# 下一层 存到  双端 队列 后面if cur.left:dq.append(cur.left)if cur.right:dq.append(cur.right)ans.append(level)return ans

方法二： 深度优先搜索 (DFS) ⟮ O ( n ) ⟯ \lgroup O(n) \rgroup ⟮O(n)⟯

参考链接

DFS 做本题的主要问题是： DFS 不是按照层次遍历的。为了让递归的过程中同一层的节点放到同一个列表中，在递归时要记录每个节点的深度。递归到新节点要把该节点 对应深度列表的末尾。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]:# 子模块def helper(node, depth):# 记住结点的 深度if not node: return if len(res) == depth:res.append([])res[depth].append(node.val)if node.left: helper(node.left, depth+1)if node.right: helper(node.right, depth+1)# 主模块if not root: return []res = []helper(root, 0)return res 

C++

方法一： 广度优先搜索 (BFS) ⟮ O ( n ) ⟯ \lgroup O(n) \rgroup ⟮O(n)⟯

/*** Definition for a binary tree node.* struct TreeNode {*     int val;*     TreeNode *left;*     TreeNode *right;*     TreeNode() : val(0), left(nullptr), right(nullptr) {}*     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}*     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}* };*/
class Solution {
public:vector<vector<int>> levelOrder(TreeNode* root) {vector<vector<int>> res;if (root == nullptr){return res;}queue <TreeNode*> q;q.emplace(root);while (!q.empty()){int n = q.size();res.emplace_back(vector<int> ());// 提前添加 空容器for (int i = 0; i < n; ++i){auto node = q.front(); q.pop();res.back().emplace_back(node->val);if (node->left){q.emplace(node->left);}if (node->right){q.emplace(node->right);}}}return res;}
};