How do you solve Number of Good Leaf Nodes Pairs on LeetCode?

Number of Good Leaf Nodes Pairs (LeetCode #1530) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n) time complexity and O(n) space complexity. Key insight: Leaf nodes are the only nodes we care about for counting pairs.

What is the brute force solution for Number of Good Leaf Nodes Pairs?

The brute force approach for Number of Good Leaf Nodes Pairs is Brute Force, with O(n²) time complexity and O(1) space complexity. The brute force approach involves finding all pairs of leaf nodes and calculating the distance between each pair. If the distance is less than or equal to the given threshold, we count that pair as 'good'. This method is straightforward but inefficient. The optimal Optimal Solution approach improves this to O(n).

What algorithm or data structure is used to solve Number of Good Leaf Nodes Pairs?

Number of Good Leaf Nodes Pairs uses the following concepts: Tree, Depth-First Search, Binary Tree. The recommended patterns to study are: Depth-First Search, Dynamic Programming.

What are the interview tips for Number of Good Leaf Nodes Pairs?

Think about how to reduce the number of calculations needed. Practice DFS problems to become comfortable with tree traversals. Always clarify the problem constraints before jumping into coding.

What are common mistakes when solving Number of Good Leaf Nodes Pairs?

Not considering the case where the tree has only one leaf. Failing to correctly implement the distance calculation.

#1530

Number of Good Leaf Nodes Pairs

Medium

TreeDepth-First SearchBinary TreeDepth-First SearchDynamic Programming

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n)
Space	O(1)	O(n)

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Intuition

Time O(n)Space O(n)

The optimal approach uses Depth-First Search (DFS) to count pairs of leaf nodes efficiently. Instead of calculating distances between all pairs, we keep track of the number of leaf nodes at each distance from the current node, allowing us to count good pairs on-the-fly.

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Algorithm

3 steps

1Step 1: Perform a DFS traversal of the tree, keeping track of the number of leaf nodes at each distance from the current node.
2Step 2: When reaching a leaf node, return an array that counts how many leaf nodes exist at each distance from this node.
3Step 3: For each pair of leaf nodes found at the current node, check if their combined distance is within the limit and count them.

solution.py27 lines

1# Full working Python code
2class TreeNode:
3    def __init__(self, val=0, left=None, right=None):
4        self.val = val
5        self.left = left
6        self.right = right
7
8class Solution:
9    def countPairs(self, root: TreeNode, distance: int) -> int:
10        self.count = 0
11        self.dfs(root, distance)
12        return self.count
13
14    def dfs(self, node, distance):
15        if not node:
16            return [0] * (distance + 1)
17        if not node.left and not node.right:
18            return [1] + [0] * distance
19        left_counts = self.dfs(node.left, distance)
20        right_counts = self.dfs(node.right, distance)
21        for i in range(distance):
22            self.count += left_counts[i] * right_counts[distance - 2 - i]
23        counts = [0] * (distance + 1)
24        for i in range(distance):
25            counts[i + 1] = left_counts[i] + right_counts[i]
26        return counts
27

ℹ

Complexity note: The complexity is O(n) because we traverse each node once, and the space complexity is O(n) due to the recursion stack and the array used to count distances.

1Leaf nodes are the only nodes we care about for counting pairs.
2Using DFS allows us to efficiently calculate distances without redundant calculations.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.