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3319. K-th Largest Perfect Subtree Size in Binary Tree

SourceWeekly Contest 419 Q2DifficultyMediumRating1603

Description

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.

 

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 2nd 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

Solutions

Solution 1: DFS + Sorting

Thinking

A perfect binary subtree needs two perfect children of equal size. With \(n \le 2000\), one post-order walk can collect every valid size.

An empty tree has size \(0\). If both children return the same non-negative size, the current tree is perfect and we record \(l+r+1\); otherwise we return \(-1\).

After sorting the sizes descending, we return the \(k\)-th or \(-1\) if fewer than \(k\) exist.

We define a function \(\textit{dfs}\) to calculate the size of the perfect binary subtree rooted at the current node, using an array \(\textit{nums}\) to record the sizes of all perfect binary subtrees. If the subtree rooted at the current node is not a perfect binary subtree, it returns \(-1\).

The execution process of the function \(\textit{dfs}\) is as follows:

  1. If the current node is null, return \(0\);
  2. Recursively calculate the sizes of the perfect binary subtrees of the left and right subtrees, denoted as \(l\) and \(r\) respectively;
  3. If the sizes of the left and right subtrees are not equal, or if the sizes of the left and right subtrees are less than \(0\), return \(-1\);
  4. Calculate the size of the perfect binary subtree rooted at the current node \(\textit{cnt} = l + r + 1\), and add \(\textit{cnt}\) to the array \(\textit{nums}\);
  5. Return \(\textit{cnt}\).

We call the \(\textit{dfs}\) function to calculate the sizes of all perfect binary subtrees. If the length of the array \(\textit{nums}\) is less than \(k\), return \(-1\). Otherwise, sort the array \(\textit{nums}\) in descending order and return the \(k\)-th largest perfect binary subtree size.

The time complexity is \(O(n \times \log n)\), and the space complexity is \(O(n)\). Here, \(n\) is the number of nodes in the binary tree.

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# 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]

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/**
 * 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;
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int kthLargestPerfectSubtree(TreeNode* root, int k) {
        vector<int> nums;
        auto dfs = [&](this auto&& dfs, TreeNode* root) -> int {
            if (!root) {
                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.push_back(cnt);
            return cnt;
        };
        dfs(root);
        if (nums.size() < k) {
            return -1;
        }
        ranges::sort(nums, greater<int>());
        return nums[k - 1];
    }
};

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/**
 * 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]
}

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/**
 * 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];
}

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