Number of Good Leaf Nodes Pairs
Difficulty:Medium
Problem
Given the root of a binary tree and an integer distance. A pair of two different leaf nodes of a binary tree is said to be good if the length of the shortest path between them is less than or equal to distance.
Return the number of good leaf node pairs in the tree.
Example 1:
Input: root = [1,2,3,null,4], distance = 3
Output: 1
Explanation: The leaf nodes of the tree are 3 and 4 and the length of the shortest path between them is 3. This is the only good pair.
Example 2:
Input: root = [1,2,3,4,5,6,7], distance = 3
Output: 2
Explanation: The good pairs are [4,5] and [6,7] with shortest path = 2. The pair [4,6] is not good because the length of ther shortest path between them is 4.
Example 3:
Input: root = [7,1,4,6,null,5,3,null,null,null,null,null,2], distance = 3
Output: 1
Explanation: The only good pair is [2,5].
Example 4:
Input: root = [100], distance = 1
Output: 0
Constraints:
The number of nodes in the tree is in the range [1, 2^10].
Each node's value is between [1, 100].
1 <= distance <= 10
Recommended: Please try your approach on your integrated development environment (IDE) first, before moving on to the solution.
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Solution:
( Java )
class Solution {
int ans = 0;
public int countPairs(TreeNode root, int distance) {
ans = 0;
dfs(root, distance);
return ans;
}
List<Integer> dfs(TreeNode root, int distance)
{
if(root == null)return new ArrayList<Integer>();
if(root.left == null && root.right == null){
List<Integer> ret = new ArrayList<>();
ret.add(0);
return ret;
}
List<Integer> L = dfs(root.left, distance);
List<Integer> R = dfs(root.right, distance);
for(int x : L){
for(int y : R){
if(x + y + 2 <= distance){
ans++;
}
}
}
List<Integer> ret = new ArrayList<>();
for(int x : L)ret.add(x+1);
for(int x : R)ret.add(x+1);
return ret;
}
}
Solution:
( Python )
class Solution(object):
def countPairs(self, root, distance):
graph = collections.defaultdict(list)
M = {}
nodes = []
leaves = set()
self.t = 0
def dfs(node):
if node:
M[node] = ans = self.t
self.t += 1
v1 = dfs(node.left)
v2 = dfs(node.right)
if v1 is not None:
graph[ans].append(v1)
graph[v1].append(ans)
if v2 is not None:
graph[ans].append(v2)
graph[v2].append(ans)
if v1 is v2 is None:
leaves.add(ans)
return ans
dfs(root)
Minv = [0] * len(M)
for node in M:
i = M[node]
Minv[i] = node
#print("!", leaves, graph)
ans = 0
for i in leaves:
rooty = i
queue = [[i, 0]]
seen = {i}
for node, d in queue:
if d > distance: break
if node in leaves and node != rooty:
#print("!", node, d)
ans += 1
for nei in graph[node]:
if nei not in seen:
seen.add(nei)
queue.append([nei, d + 1])
return ans // 2
If you have any doubts regarding this problem or need the solution in other programming languages then leave a comment down below .
