Question

2792. Count Nodes That Are Great Enough

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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

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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) }