O(N) DFS + Bucket Sort
ApproachThe number of nodes in the tree N is bounded from above by 2000. This means the largest perfect subtree will never exceed 211=2048 nodes.The number of n
# Approach
The number of nodes in the tree $$N$$ is bounded from above by 2000. This means the largest perfect subtree will never exceed $$2^{11} = 2048$$ nodes. 

The number of nodes in a perfect subtree is $$1 + 2 + 4 + ... + 2^m = 2^{m + 1} - 1$$. The binary representation of this number is all 1s. We can use an array to hold the counts of perfect subtrees we encounter. 

Below, it is called `bitcounts`. In particular, `bitcounts[j]` is the number of perfect subtrees with $$2^j - 1$$ nodes. 

We use DFS in the method `isPerfect(node: Optional[TreeNode]) -> tuple[bool, int]`. The method returns `True` and and the number of nodes for the root of a perfect subtree and `False, 0` otherwise. At each call of the method, whenever we see a perfect subtree with $$2^j - 1$$ nodes, we'll increment our counter at `bitcounts[j]`. 

This bucket sorts the counts of perfect subtrees by the bitcount of the number of nodes in the perfect subtree. 

We can then read `bitcounts` from right to left, decrementing `k` by the count of perfect subtrees until `k <= 0`. At that point, we have found the number of nodes in `kth` largest perfect subtree, and we can return the desired number. Otherwise we return `-1`. 

# Complexity
- Time complexity:
<!-- Add your time complexity here, e.g. $$O(n)$$ -->

- Space complexity:
<!-- Add your space complexity here, e.g. $$O(n)$$ -->

# Code
```python3 []
# 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 kthLargestPerfectSubtree(self, root: Optional[TreeNode], k: int) -> int:
        bitcounts = [0] * 11

        def isPerfect(node: Optional[TreeNode]) -> tuple[bool, int]:
            if not node:
                return True, 0
            
            leftPerfect, leftCount = isPerfect(node.left)
            rightPerfect, rightCount = isPerfect(node.right)

            if (leftPerfect and rightPerfect) and (leftCount == rightCount):
                # number of nodes in the perfect subtree
                size = 2 * left[1] + 1  

                # number of set bits in the size of the perfect subtree
                bits = size.bit_count()

                # update bucket sort array with frequency
                bitcounts[bits] += 1
                return True, size
            else:
                return False, 0
        
        _ = isPerfect(root)
        for j in range(len(bitcounts) - 1, -1, -1):
            k -= bitcounts[j]
            if k <= 0:
                return 2**j - 1
        return -1
```
['Python3']