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2792. Count Nodes That Are Great Enough πŸ”’

Description

You are given a root to a binary tree and an integer k. A node of this tree is called great enough if the followings hold:

  • Its subtree has at least k nodes.
  • Its value is greater than the value of at least k nodes in its subtree.

Return the number of nodes in this tree that are great enough.

The node u is in the subtree of the node v, if u == v or v is an ancestor of u.

 

Example 1:

Input: root = [7,6,5,4,3,2,1], k = 2
Output: 3
Explanation: Number the nodes from 1 to 7.
The values in the subtree of node 1: {1,2,3,4,5,6,7}. Since node.val == 7, there are 6 nodes having a smaller value than its value. So it's great enough.
The values in the subtree of node 2: {3,4,6}. Since node.val == 6, there are 2 nodes having a smaller value than its value. So it's great enough.
The values in the subtree of node 3: {1,2,5}. Since node.val == 5, there are 2 nodes having a smaller value than its value. So it's great enough.
It can be shown that other nodes are not great enough.
See the picture below for a better understanding.

Example 2:

Input: root = [1,2,3], k = 1
Output: 0
Explanation: Number the nodes from 1 to 3.
The values in the subtree of node 1: {1,2,3}. Since node.val == 1, there are no nodes having a smaller value than its value. So it's not great enough.
The values in the subtree of node 2: {2}. Since node.val == 2, there are no nodes having a smaller value than its value. So it's not great enough.
The values in the subtree of node 3: {3}. Since node.val == 3, there are no nodes having a smaller value than its value. So it's not great enough.
See the picture below for a better understanding.

Example 3:

Input: root = [3,2,2], k = 2
Output: 1
Explanation: Number the nodes from 1 to 3.
The values in the subtree of node 1: {2,2,3}. Since node.val == 3, there are 2 nodes having a smaller value than its value. So it's great enough.
The values in the subtree of node 2: {2}. Since node.val == 2, there are no nodes having a smaller value than its value. So it's not great enough.
The values in the subtree of node 3: {2}. Since node.val == 2, there are no nodes having a smaller value than its value. So it's not great enough.
See the picture below for a better understanding.

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104]. 
  • 1 <= Node.val <= 104
  • 1 <= k <= 10

Solutions

Solution 1

Thinking

A node is great enough iff its subtree contains at least \(k\) strictly smaller values. Sorting every subtree wastes work when \(k\) is small.

A post-order DFS merges at most \(k\) smallest child values in a heap of size \(k\). If the heap is full and its top (the negated \(k\)-th smallest) is still less than the node, the node is counted.

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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 countGreatEnoughNodes(self, root: Optional[TreeNode], k: int) -> int:
        def push(pq, x):
            heappush(pq, x)
            if len(pq) > k:
                heappop(pq)

        def dfs(root):
            if root is None:
                return []
            l, r = dfs(root.left), dfs(root.right)
            for x in r:
                push(l, x)
            if len(l) == k and -l[0] < root.val:
                nonlocal ans
                ans += 1
            push(l, -root.val)
            return l

        ans = 0
        dfs(root)
        return ans

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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 int ans;
    private int k;

    public int countGreatEnoughNodes(TreeNode root, int k) {
        this.k = k;
        dfs(root);
        return ans;
    }

    private PriorityQueue<Integer> dfs(TreeNode root) {
        if (root == null) {
            return new PriorityQueue<>(Comparator.reverseOrder());
        }
        var l = dfs(root.left);
        var r = dfs(root.right);
        for (int x : r) {
            l.offer(x);
            if (l.size() > k) {
                l.poll();
            }
        }
        if (l.size() == k && l.peek() < root.val) {
            ++ans;
        }
        l.offer(root.val);
        if (l.size() > k) {
            l.poll();
        }
        return l;
    }
}

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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 countGreatEnoughNodes(TreeNode* root, int k) {
        int ans = 0;
        function<priority_queue<int>(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return priority_queue<int>();
            }
            auto left = dfs(root->left);
            auto right = dfs(root->right);
            while (right.size()) {
                left.push(right.top());
                right.pop();
                if (left.size() > k) {
                    left.pop();
                }
            }
            if (left.size() == k && left.top() < root->val) {
                ++ans;
            }
            left.push(root->val);
            if (left.size() > k) {
                left.pop();
            }
            return left;
        };
        dfs(root);
        return ans;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func countGreatEnoughNodes(root *TreeNode, k int) (ans int) {
    var dfs func(*TreeNode) hp
    dfs = func(root *TreeNode) hp {
        if root == nil {
            return hp{}
        }
        l, r := dfs(root.Left), dfs(root.Right)
        for _, x := range r.IntSlice {
            l.push(x)
            if l.Len() > k {
                l.pop()
            }
        }
        if l.Len() == k && root.Val > l.IntSlice[0] {
            ans++
        }
        l.push(root.Val)
        if l.Len() > k {
            l.pop()
        }
        return l
    }
    dfs(root)
    return
}

type hp struct{ sort.IntSlice }

func (h hp) Less(i, j int) bool { return h.IntSlice[i] > h.IntSlice[j] }
func (h *hp) Push(v any)        { h.IntSlice = append(h.IntSlice, v.(int)) }
func (h *hp) Pop() any {
    a := h.IntSlice
    v := a[len(a)-1]
    h.IntSlice = a[:len(a)-1]
    return v
}
func (h *hp) push(v int) { heap.Push(h, v) }
func (h *hp) pop() int   { return heap.Pop(h).(int) }

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