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2792. Count Nodes That Are Great Enough

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

  • /**
     * 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.
    # 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.
     * 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 interface{}) { h.IntSlice = append(h.IntSlice, v.(int)) }
    func (h *hp) Pop() interface{} {
    	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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