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  • 【leetcode】662. Maximum Width of Binary Tree

    题目如下:

    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.

    解题思路:对于一个二叉树,我们可以按层序遍历的顺序给每一个节点定义一个顺序索引,例如根节点是第一个遍历的节点,那么索引是1。很显然,根节点的左节点的索引是2,右节点是3。二叉树的父节点与左右子节点的索引满足这么一个规律的,如果父节点的索引值是i,那么左节点是2*i,右节点是2*i+1。所以,只需要用遍历二叉树,计算出每一层最左边的节点和最右边节点的索引的差值即可。

    代码如下:

    # 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):
        dic = {}
        res = 0
        def traverse(self,node,level,inx):
            if node == None:
                return 
            if level not in self.dic:
                self.dic[level] = [inx]
            else:
                if len(self.dic[level]) == 1:
                    self.dic[level].append(inx)
                else:
                    self.dic[level][1] = inx
                self.res = max(self.res,self.dic[level][1] - self.dic[level][0])
            if node.left != None:
                self.traverse(node.left,level + 1 ,inx*2)
            if node.right != None:
                self.traverse(node.right, level + 1, inx * 2 + 1)
    
        def widthOfBinaryTree(self, root):
            """
            :type root: TreeNode
            :rtype: int
            """
            self.dic = {}
            self.res = 0
            self.traverse(root,0,1)
            return self.res + 1
            
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  • 原文地址:https://www.cnblogs.com/seyjs/p/10001975.html
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