［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); } }