Given the root of a binary tree, return all duplicate subtrees.

For each kind of duplicate subtrees, you only need to return the root node of any one of them.

Two trees are duplicate if they have the same structure with the same node values.

 

Example 1:

Input: root = [1,2,3,4,null,2,4,null,null,4]
Output: [[2,4],[4]]

Example 2:

Input: root = [2,1,1]
Output: [[1]]

Example 3:

Input: root = [2,2,2,3,null,3,null]
Output: [[2,3],[3]]

 

Constraints:

  • The number of the nodes in the tree will be in the range [1, 10^4]
  • -200 <= Node.val <= 200




 # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findDuplicateSubtrees(self, root: Optional[TreeNode]) -> List[Optional[TreeNode]]:

        ans = []
        hashmap = {}

        def find(root, path):
            if not root:
                return '#'

            path += ','.join([str(root.val), find(root.left, path), find(root.right, path)])

            if path in hashmap:
                hashmap[path]+=1
                if hashmap[path] == 2:
                    ans.append(root)
            else:
                hashmap[path] = 1

            return path


        find(root, "")
        return ans

Random Note


(k := next(iter(d)), d.pop(k)) will remove the leftmost (first) item (if it exists) from a dict object. And if you want to remove the right most/recent value from the dict

d.popitem()