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This approach involves a DFS traversal of the tree. For each node, we compute the number of leaf nodes at each possible distance. We merge the results from the left and right subtrees and count the valid leaf pairs.
Time Complexity: O(n * distance) due to traversing the entire tree and computing leaf pairs.
Space Complexity: O(distance) for storing distance arrays.
1class TreeNode:
2 def __init__(self, x):
3 self.val = x
4 self.left = None
5 self.right = None
6
7class Solution:
8 def countPairs(self, root: TreeNode, distance: int) -> int:
9 def dfs(node):
10 if not node:
11 return [0] * (distance + 1)
12 if not node.left and not node.right:
13 leaves = [0] * (distance + 1)
14 leaves[1] = 1
15 return leaves
16
17 left = dfs(node.left)
18 right = dfs(node.right)
19
20 for i in range(1, distance + 1):
21 for j in range(1, distance - i + 1):
22 self.result += left[i] * right[j]
23
24 leaves = [0] * (distance + 1)
25 for i in range(distance):
26 leaves[i + 1] = left[i] + right[i]
27 return leaves
28
29 self.result = 0
30 dfs(root)
31 return self.resultPython leverages a similar approach to handle distances with lists. Recursion helps traverse from leaves upwards, combining results facilitated by helper lists for each node.
This approach leverages preprocessing leaf distances in a dynamic programming table, which gives quick lookups during pair evaluation. It builds the DP table during a DFS traversal, allowing efficient pair computation afterward.
Time Complexity: Approximately O(n^2) for thorough pairing and evaluation.
Space Complexity: O(n^2) due to pair-wise distance visualization.
1/* This solution is based on C and would implement a similar strategy to preprocess and calculate distances between all leaves, followed by evaluating valid pairs. For brevity, a sample plan is provided without full implementation due to competition space. */A solution in C would involve more detailed memory management and array calculations for storing and evaluating distances.