LeetCode #3319 — MEDIUM

K-th Largest Perfect Subtree Size in Binary Tree

Move from brute-force thinking to an efficient approach using tree strategy.

Solve on LeetCode

The Problem

Problem Statement

You are given the root of a binary tree and an integer k.

Return an integer denoting the size of the k^th largest perfect binary subtree, or -1 if it doesn't exist.

A perfect binary tree is a tree where all leaves are on the same level, and every parent has two children.

Example 1:

Input: root = [5,3,6,5,2,5,7,1,8,null,null,6,8], k = 2

Output: 3

Explanation:

The roots of the perfect binary subtrees are highlighted in black. Their sizes, in non-increasing order are [3, 3, 1, 1, 1, 1, 1, 1].
The 2^nd largest size is 3.

Example 2:

Input: root = [1,2,3,4,5,6,7], k = 1

Output: 7

Explanation:

The sizes of the perfect binary subtrees in non-increasing order are [7, 3, 3, 1, 1, 1, 1]. The size of the largest perfect binary subtree is 7.

Example 3:

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

Output: -1

Explanation:

The sizes of the perfect binary subtrees in non-increasing order are [1, 1]. There are fewer than 3 perfect binary subtrees.

Constraints:

The number of nodes in the tree is in the range [1, 2000].
1 <= Node.val <= 2000
1 <= k <= 1024

Step 01

Brute Force Baseline

Problem summary: You are given the root of a binary tree and an integer k. Return an integer denoting the size of the kth largest perfect binary subtree, or -1 if it doesn't exist. A perfect binary tree is a tree where all leaves are on the same level, and every parent has two children.

Baseline thinking

Start with the most direct exhaustive search. That gives a correctness anchor before optimizing.

Pattern signal: Tree

Example 1

[5,3,6,5,2,5,7,1,8,null,null,6,8]
2

Example 2

[1,2,3,4,5,6,7]
1

Example 3

[1,2,3,null,4]
3

Core Insight

What unlocks the optimal approach

For a subtree to form a perfect binary subtree, its children should also be perfect binary subtrees.
Check recursively that both the node and its children are perfect binary subtrees.
Gather all the perfect binary subtrees and return the kth largest.

Interview move: turn each hint into an invariant you can check after every iteration/recursion step.

Step 03

Algorithm Walkthrough

Iteration Checklist

Define state (indices, window, stack, map, DP cell, or recursion frame).
Apply one transition step and update the invariant.
Record answer candidate when condition is met.
Continue until all input is consumed.

Use the first example testcase as your mental trace to verify each transition.

Step 04

Edge Cases

Minimum Input

Single element / shortest valid input

Validate boundary behavior before entering the main loop or recursion.

Duplicates & Repeats

Repeated values / repeated states

Decide whether duplicates should be merged, skipped, or counted explicitly.

Extreme Constraints

Upper-end input sizes

Re-check complexity target against constraints to avoid time-limit issues.

Invalid / Corner Shape

Empty collections, zeros, or disconnected structures

Handle special-case structure before the core algorithm path.

Step 05

Full Annotated Code

Source-backed implementations are provided below for direct study and interview prep.

// Accepted solution for LeetCode #3319: K-th Largest Perfect Subtree Size in Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private List<Integer> nums = new ArrayList<>();

    public int kthLargestPerfectSubtree(TreeNode root, int k) {
        dfs(root);
        if (nums.size() < k) {
            return -1;
        }
        nums.sort(Comparator.reverseOrder());
        return nums.get(k - 1);
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int l = dfs(root.left);
        int r = dfs(root.right);
        if (l < 0 || l != r) {
            return -1;
        }
        int cnt = l + r + 1;
        nums.add(cnt);
        return cnt;
    }
}

// Accepted solution for LeetCode #3319: K-th Largest Perfect Subtree Size in Binary Tree
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func kthLargestPerfectSubtree(root *TreeNode, k int) int {
	nums := []int{}
	var dfs func(*TreeNode) int
	dfs = func(root *TreeNode) int {
		if root == nil {
			return 0
		}
		l, r := dfs(root.Left), dfs(root.Right)
		if l < 0 || l != r {
			return -1
		}
		cnt := l + r + 1
		nums = append(nums, cnt)
		return cnt
	}
	dfs(root)
	if len(nums) < k {
		return -1
	}
	sort.Sort(sort.Reverse(sort.IntSlice(nums)))
	return nums[k-1]
}

# Accepted solution for LeetCode #3319: K-th Largest Perfect Subtree Size in Binary Tree
# 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:
        def dfs(root: Optional[TreeNode]) -> int:
            if root is None:
                return 0
            l, r = dfs(root.left), dfs(root.right)
            if l < 0 or l != r:
                return -1
            cnt = l + r + 1
            nums.append(cnt)
            return cnt

        nums = []
        dfs(root)
        if len(nums) < k:
            return -1
        nums.sort(reverse=True)
        return nums[k - 1]

// Accepted solution for LeetCode #3319: K-th Largest Perfect Subtree Size in Binary Tree
// Rust example auto-generated from java reference.
// Replace the signature and local types with the exact LeetCode harness for this problem.
impl Solution {
    pub fn rust_example() {
        // Port the logic from the reference block below.
    }
}

// Reference (java):
// // Accepted solution for LeetCode #3319: K-th Largest Perfect Subtree Size in Binary Tree
// /**
//  * Definition for a binary tree node.
//  * public class TreeNode {
//  *     int val;
//  *     TreeNode left;
//  *     TreeNode right;
//  *     TreeNode() {}
//  *     TreeNode(int val) { this.val = val; }
//  *     TreeNode(int val, TreeNode left, TreeNode right) {
//  *         this.val = val;
//  *         this.left = left;
//  *         this.right = right;
//  *     }
//  * }
//  */
// class Solution {
//     private List<Integer> nums = new ArrayList<>();
// 
//     public int kthLargestPerfectSubtree(TreeNode root, int k) {
//         dfs(root);
//         if (nums.size() < k) {
//             return -1;
//         }
//         nums.sort(Comparator.reverseOrder());
//         return nums.get(k - 1);
//     }
// 
//     private int dfs(TreeNode root) {
//         if (root == null) {
//             return 0;
//         }
//         int l = dfs(root.left);
//         int r = dfs(root.right);
//         if (l < 0 || l != r) {
//             return -1;
//         }
//         int cnt = l + r + 1;
//         nums.add(cnt);
//         return cnt;
//     }
// }

// Accepted solution for LeetCode #3319: K-th Largest Perfect Subtree Size in Binary Tree
/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function kthLargestPerfectSubtree(root: TreeNode | null, k: number): number {
    const nums: number[] = [];
    const dfs = (root: TreeNode | null): number => {
        if (!root) {
            return 0;
        }
        const l = dfs(root.left);
        const r = dfs(root.right);
        if (l < 0 || l !== r) {
            return -1;
        }
        const cnt = l + r + 1;
        nums.push(cnt);
        return cnt;
    };
    dfs(root);
    if (nums.length < k) {
        return -1;
    }
    return nums.sort((a, b) => b - a)[k - 1];
}

Step 06

Interactive Study Demo

Use this to step through a reusable interview workflow for this problem.

Custom checkpoints (one per line)

Press Step or Run All to begin.

Step 07

Complexity Analysis

Time

O(n × log n)

Space

O(n)

Approach Breakdown

LEVEL ORDER

O(n) time

O(n) space

BFS with a queue visits every node exactly once — O(n) time. The queue may hold an entire level of the tree, which for a complete binary tree is up to n/2 nodes = O(n) space. This is optimal in time but costly in space for wide trees.

DFS TRAVERSAL

O(n) time

O(h) space

Every node is visited exactly once, giving O(n) time. Space depends on tree shape: O(h) for recursive DFS (stack depth = height h), or O(w) for BFS (queue width = widest level). For balanced trees h = log n; for skewed trees h = n.

Shortcut: Visit every node once → O(n) time. Recursion depth = tree height → O(h) space.

Coach Notes

Common Mistakes

Review these before coding to avoid predictable interview regressions.

Forgetting null/base-case handling

Wrong move: Recursive traversal assumes children always exist.

Usually fails on: Leaf nodes throw errors or create wrong depth/path values.

Fix: Handle null/base cases before recursive transitions.

K-th Largest Perfect Subtree Size in Binary Tree

Problem Statement

Roadmap

Brute Force Baseline

Baseline thinking

Example 1

Example 2

Example 3

Related Problems

Core Insight

What unlocks the optimal approach

Algorithm Walkthrough

Iteration Checklist

Edge Cases

Full Annotated Code

Interactive Study Demo

Complexity Analysis

Approach Breakdown

Common Mistakes

Forgetting null/base-case handling