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

［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] ：

/**
 * 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);
    }
}