码迷,mamicode.com
首页 > 其他好文 > 详细

662. Maximum Width of Binary Tree二叉树的最大宽度

时间:2018-08-28 10:38:40      阅读:254      评论:0      收藏:0      [点我收藏+]

标签:spl   length   new   span   初始   tin   cond   space   get   

[抄题]:

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: 

          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).

 [暴力解法]:

时间分析:

空间分析:

 [优化后]:

时间分析:

空间分析:

[奇葩输出条件]:

[奇葩corner case]:

从第三层开始,如果已经满了,就要添加新的index

if (level == list.size()) list.add(index);

 

[思维问题]:

完全没思路,因此需要一些基础知识

[英文数据结构或算法,为什么不用别的数据结构或算法]:

We know that a binary tree can be represented by an array (assume the root begins from the position with index 1 in the array). If the index of a node is i, the indices of its two children are 2*i and 2*i + 1. The idea is to use two arrays (start[] and end[]) to record the the indices of the leftmost node and rightmost node in each level, respectively. For each level of the tree, the width isend[level] - start[level] + 1. Then, we just need to find the maximum width.

[一句话思路]:

[输入量]:空: 正常情况:特大:特小:程序里处理到的特殊情况:异常情况(不合法不合理的输入):

[画图]:

[一刷]:

index的初始值为啥是1?不懂,算了

[二刷]:

[三刷]:

[四刷]:

[五刷]:

  [五分钟肉眼debug的结果]:

[总结]:

找参数只是结果,重要的是把所需的变量找出来

还是按照起点、过程、终点来写,index的左右分别为 2 * index 和  2 * index + 1,

[复杂度]:Time complexity: O(n) Space complexity: O(n)

[算法思想:迭代/递归/分治/贪心]:

[关键模板化代码]:

[其他解法]:

[Follow Up]:

[LC给出的题目变变变]:

 [代码风格] :

 [是否头一次写此类driver funcion的代码] :

 [潜台词] :

 

技术分享图片
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int max = 1;
    
    public int widthOfBinaryTree(TreeNode root) {
        //corner case
        if (root == null) return 0;
        
        //initialization
        List<Integer> startOfLevel = new ArrayList<Integer>();
        
        //return
        getWidth(root, 1, 0, startOfLevel);
        return max;
    }
    
    public void getWidth(TreeNode root, int index, int level, List<Integer> list) {
        //return null
        if (root == null) return ;
        
        //add the index to list
        if (list.size() == level)
            list.add(index);
        
        max = Math.max(max, index + 1 - list.get(level));
        
        //divide and conquer in left and right
        getWidth(root.left, 2 * index, level + 1, list);
        getWidth(root.right, 2 * index + 1, level + 1, list);
    }
}
View Code

 

662. Maximum Width of Binary Tree二叉树的最大宽度

标签:spl   length   new   span   初始   tin   cond   space   get   

原文地址:https://www.cnblogs.com/immiao0319/p/9545768.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!