Leaf Pairs Less Than Target Distance Away - Google Top Interview Questions
Problem Statement :
You are given a binary tree root and an integer target. Return the number of pairs of leaves such that the shortest distance between them is less than or equal to target. Constraints 0 ≤ n, target ≤ 1,000 where n is the number of nodes in root Example 1 Input root = [1, [2, [4, null, null], [5, null, null]], [3, null, null]] target = 2 Output 1 Explanation The pair (4, 5) meet the distance criteria since the distance between them is 2. Example 2 Input root = [1, [2, [4, null, null], [5, null, null]], [3, null, null]] target = 4 Output 3 Explanation Now pairs (4, 5), (3, 4) and (3, 5) meet the distance criteria.
Solution :
Solution in C++ :
vector<int> tra(Tree* root, int k, int& ans) {
if (!root) return {};
if (root->left == root->right) {
return {0, 1};
}
vector<int> left = tra(root->left, k, ans);
vector<int> right = tra(root->right, k, ans);
for (int i = 1; i < left.size(); i++) {
for (int j = 1; j < right.size(); j++) {
if (i + j <= k) {
ans += left[i] * right[j];
}
}
}
int final_len = max((int)left.size(), (int)right.size()) + 1;
vector<int> ret(final_len + 1, 0);
for (int i = 2; i < final_len; i++) {
if (left.size() >= i) ret[i] += left[i - 1];
if (right.size() >= i) ret[i] += right[i - 1];
}
return ret;
}
int solve(Tree* root, int target) {
int ans = 0;
tra(root, target, ans);
return ans;
}
Solution in Java :
import java.util.*;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
HashMap<Tree, ArrayList<Tree>> graph;
ArrayList<Tree> leaves;
public boolean is_leaf(Tree root) {
return root.left == null && root.right == null;
}
public int solve(Tree root, int target) {
graph = new HashMap();
leaves = new ArrayList();
fill(root);
HashSet<Pair<Tree, Tree>> hs = new HashSet();
for (Tree leaf : leaves) {
int dist = 0;
LinkedList<Tree> q = new LinkedList();
HashSet<Tree> visited = new HashSet();
visited.add(leaf);
q.add(leaf);
while (q.size() > 0 && dist < target) {
int size = q.size();
for (int i = 0; i < size; i++) {
Tree cur = q.removeFirst();
visited.add(cur);
for (Tree neighbor : graph.get(cur)) {
if (visited.contains(neighbor))
continue;
if (is_leaf(neighbor))
hs.add(new Pair(neighbor, leaf));
q.addLast(neighbor);
visited.add(neighbor);
}
}
dist++;
}
}
return hs.size() / 2;
}
public void fill(Tree root) {
if (root == null)
return;
if (root.left == null && root.right == null) {
leaves.add(root);
}
graph.putIfAbsent(root, new ArrayList());
if (root.left != null) {
graph.get(root).add(root.left);
graph.putIfAbsent(root.left, new ArrayList());
graph.get(root.left).add(root);
}
if (root.right != null) {
graph.get(root).add(root.right);
graph.putIfAbsent(root.right, new ArrayList());
graph.get(root.right).add(root);
}
fill(root.left);
fill(root.right);
}
}
Solution in Python :
# class Tree:
# def __init__(self, val, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def solve(self, root, target):
res = 0
def dfs(root):
nonlocal res
if not root:
return []
if root and not root.left and not root.right:
return [1]
left, right = dfs(root.left), dfs(root.right)
right.sort()
for x in left:
res += bisect.bisect(right, target - x)
return [x + 1 for x in left] + [x + 1 for x in right]
dfs(root)
return res
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