LeetCode | 0102. Binary Tree Level Order Traversal二叉树的层次遍历【Python】

小咪咪 2023-07-17 11:20 82阅读 0赞

> LeetCode 0102. Binary Tree Level Order Traversal二叉树的层次遍历【Medium】【Python】【BFS】

### Problem ###

[LeetCode][]

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,null,null,15,7]`,

3
       / \
      9  20
        /  \
       15   7

return its level order traversal as:

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

### 问题 ###

[力扣][Link 1]

给定一个二叉树，返回其按层次遍历的节点值。 （即逐层地，从左到右访问所有节点）。

例如:  
给定二叉树: `[3,9,20,null,null,15,7]`,

3
       / \
      9  20
        /  \
       15   7

返回其层次遍历结果：

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

### 思路 ###

**BFS**

##### 解法一：非递归 #####

当队列不为空：
    	当前层打印循环：
    		队首元素出队，记为 node
    		将 node.val 添加到 temp 尾部
    		若左（右）子节点不为空，则将左（右）子节点加入队列
    	把当前 temp 中的所有元素加入 res

**时间复杂度:** O(n)，n 为二叉树的节点数。  
**空间复杂度:** O(n)，n 为二叉树的节点数。

##### Python3代码 #####

# Definition for a binary tree node.
    # class TreeNode:
    # def __init__(self, x):
    # self.val = x
    # self.left = None
    # self.right = None
    
    class Solution:
        def levelOrder(self, root: TreeNode) -> List[List[int]]:
            # solution one: 非递归
            import collections
            if not root:
                return []
            
            res, q = [], collections.deque()
            q.append(root)
            while q:
                # 输出是二维数组
                temp = []
                for x in range(len(q)):
                    node = q.popleft()
                    temp.append(node.val)
                    if node.left:
                        q.append(node.left)
                    if node.right:
                        q.append(node.right)
                res.append(temp)
            return res

##### 解法二：递归 #####

##### Python3代码 #####

# Definition for a binary tree node.
    # class TreeNode:
    # def __init__(self, x):
    # self.val = x
    # self.left = None
    # self.right = None
    
    class Solution:
        def levelOrder(self, root: TreeNode) -> List[List[int]]:
            # solution two: 递归
            res = []
            
            def helper(root, depth):
                if not root:
                    return
                if len(res) == depth:
                    res.append([])
                res[depth].append(root.val)
                helper(root.left, depth + 1)
                helper(root.right, depth + 1)
            helper(root , 0)
            return res

### GitHub链接 ###

[Python][]

[LeetCode]: https://leetcode.com/problems/binary-tree-level-order-traversal/
[Link 1]: https://leetcode-cn.com/problems/binary-tree-level-order-traversal/
[Python]: https://github.com/Wonz5130/LeetCode-Solutions/blob/master/solutions/0102-Binary-Tree-Level-Order-Traversal/0102.py