[leetcode] 662. Maximum Width of Binary Tree @ python

最新推荐文章于 2022-04-03 22:41:31 发布

原创最新推荐文章于 2022-04-03 22:41:31 发布 · 244 阅读

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本文探讨了如何求解二叉树的最大宽度，通过两种不同的方法实现：使用deque和列表进行BFS遍历，每层节点按序号计算宽度，最终得出最大宽度。示例展示了不同二叉树结构下最大宽度的计算。

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原题

Given a binary tree, write a function to get the maximum width of the given tree. The width of a tree is the maximum width among all levels. The binary tree has the same structure as a full binary tree, but some nodes are null.

The width of one level is defined as the length between the end-nodes (the leftmost and right most non-null nodes in the level, where the null nodes between the end-nodes are also counted into the length calculation.

Example 1:
Input:

       1
     /   \
    3     2
   / \     \  
  5   3     9

Output: 4
Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).
Example 2:
Input:

Output: 2
Explanation: The maximum width existing in the third level with the length 2 (5,3).
Example 3:
Input:

Output: 2
Explanation: The maximum width existing in the second level with the length 2 (3,2).
Example 4:
Input:

      1
     / \
    3   2
   /     \  
  5       9 
 /         \
6           7

Output: 8
Explanation:The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).

解法1

我们假设树是完全二叉树, 这样可以对每个节点计数, 我们设根节点的序号为1, 在第n个节点时, 它的左子树是第2n个节点, 它的右子树是2n+1个节点, 然后使用BFS, 如果子树存在的话, 将序号放入列表中, 宽度是 = 最右边的序号 - 最左边的序号 + 1, 每次循环时更新res. 我们使用deque存放每层节点.
Time: O(n)
Space: O(n)

代码

class Solution:
    def widthOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # base case
        if not root: return 0
        q = collections.deque()
        q.append((root, 1))
        res = 0
        while q:
            width = q[-1][1] - q[0][1] + 1
            res = max(res, width)
            for i in range(len(q)):
                node, count = q.popleft()
                if node.left:
                    q.append((node.left, 2*count))
                if node.right:
                    q.append((node.right, 2*count + 1))
                    
        return res

解法2

同解法1, 使用列表存放每层的节点

代码

class Solution:
    def widthOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        # base case
        if not root: return 0
        q = [(root, 1)]
        res = 0
        while q:
            width = q[-1][1] - q[0][1] + 1
            res = max(res, width)
            new_q = []
            for tup in q:
                node, count = tup[0], tup[1]
                if node.left:
                    new_q.append((node.left, 2*count))
                if node.right:
                    new_q.append((node.right, 2*count + 1))
            q = new_q
                    
        return res