LeetCode 2792. Count Nodes That Are Great Enough Solution in Java, C++, Python & Go | Explanation + Code

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

PythonJavaC++Go

# 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(code-box)

/** * 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; } }(code-box)

/** * 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; } };(code-box)

/** * 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) }(code-box)