[LeetCode]Maximum Width of Binary Tree

题目描述:

LeetCode 662. Maximum Width of Binary Tree

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: 

          1
         /  
        3    
       / \       
      5   3     

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

Example 3:

Input: 

          1
         / \
        3   2 
       /        
      5      

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).

Note: Answer will in the range of 32-bit signed integer.

题目大意:

给定二叉树,求二叉树的最大宽度。二叉树某层的宽度是指其最左非空节点与最右非空节点之间的跨度。

解题思路:

二叉树层次遍历

对二叉树节点进行标号,根节点标号为1;若某节点标号为c,则其左右孩子标号分别为2c, 2c + 1

某层的宽度即为最右、最左节点标号之差+1

Python代码:

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def widthOfBinaryTree(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        q = [(root, 1)]
        ans = 0
        while q:
            width = q[-1][-1] - q[0][-1] + 1
            ans = max(ans, width)
            q0 = []
            for n, i in q:
                if n.left: q0.append((n.left, i * 2))
                if n.right: q0.append((n.right, i * 2 + 1))
            q = q0
        return ans

 

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