動機

忘了可以上index，要算距離就是index!!

Problem

Given the root of a binary tree, return the maximum width of the given tree.

The maximum width of a tree is the maximum width among all levels.

The width of one level is defined as the length between the end-nodes (the leftmost and rightmost non-null nodes), where the null nodes between the end-nodes are also counted into the length calculation.

It is guaranteed that the answer will in the range of 32-bit signed integer.

 

Example 1:

Input: root = [1,3,2,5,3,null,9]Output: 4Explanation: The maximum width existing in the third level with the length 4 (5,3,null,9).

Example 2:

Input: root = [1,3,null,5,3]Output: 2Explanation: The maximum width existing in the third level with the length 2 (5,3).

Example 3:

Input: root = [1,3,2,5]Output: 2Explanation: The maximum width existing in the second level with the length 2 (3,2).

Example 4:

Input: root = [1,3,2,5,null,null,9,6,null,null,7]Output: 8Explanation: The maximum width existing in the fourth level with the length 8 (6,null,null,null,null,null,null,7).

 

Constraints:

  • The number of nodes in the tree is in the range [1, 3000].
  • -100 <= Node.val <= 100

Sol

把tree每一level去編index

0
0 1
0 1 2 3

就能推出i*2與i*2+1這兩個規則

class Solution:
    def widthOfBinaryTree(self, root: TreeNode) -> int:
        starts,ret = {},1
        
        def dfs(r,level=0,i=0):
            nonlocal ret
            if not r:
                return
            else:
                if level not in starts:
                    starts[level] = i
                else:
                    ret = max(ret, i-starts[level]+1)
                dfs(r.left,level+1,i*2)
                dfs(r.right,level+1,i*2+1)
        
        dfs(root)
        return ret