Binary Tree Longest Consecutive Path - Amazon Top Interview Questions
Problem Statement :
Given a binary tree root, return the length of the longest path consisting of consecutive values between any two nodes in the tree. The path can be consecutively increasing or decreasing. Constraints n ≤ 100,000 where n is the number of nodes in root Example 1 Input root = [2, [1, null, null], [3, [4, [5, null, null], null], [0, null, null]]] Output 5 Explanation A longest path is [1, 2, 3, 4, 5]. Example 2 Input root = [7, null, [6, [8, [4, null, null], null], [5, null, [4, null, null]]]] Output 4 Explanation A longest path is [7, 6, 5, 4]. Example 3 Input root = [2, [1, [0, null, null], null], null] Output 3 Explanation A longest path is [2, 1, 0].
Solution :
Solution in C++ :
int ans;
// increasing / decreasing
pair<int, int> dfs(Tree* root) {
if (!root) return {0, 0};
int inc = 1, dec = 1;
auto l = dfs(root->left);
auto r = dfs(root->right);
if (root->left) {
if (root->val == root->left->val + 1) dec = max(dec, 1 + l.second);
if (root->val == root->left->val - 1) inc = max(inc, 1 + l.first);
}
if (root->right) {
if (root->val == root->right->val + 1) dec = max(dec, 1 + r.second);
if (root->val == root->right->val - 1) inc = max(inc, 1 + r.first);
}
ans = max(ans, inc + dec - 1);
return {inc, dec};
}
int solve(Tree* root) {
ans = 1;
dfs(root);
return ans;
}
Solution in Java :
import java.util.*;
import java.util.concurrent.atomic.AtomicInteger;
/**
* public class Tree {
* int val;
* Tree left;
* Tree right;
* }
*/
class Solution {
public int solve(Tree root) {
AtomicInteger max = new AtomicInteger(0);
traversal(root, max);
return max.get();
}
private boolean isLeafNode(Tree node) {
return node.left == null && node.right == null;
}
private int traversal(Tree node, AtomicInteger max) {
if (node == null) {
max.set(Math.max(max.get(), 0));
return 0;
} else if (isLeafNode(node)) {
max.set(Math.max(max.get(), 1));
return 1;
}
int left = traversal(node.left, max);
int right = traversal(node.right, max);
int mineWithLeft = 1;
int mineWithRight = 1;
int collective = 0;
if (left != 0 && Math.abs(node.left.val - node.val) == 1) {
mineWithLeft += left;
collective += left;
}
if (right != 0 && Math.abs(node.right.val - node.val) == 1) {
mineWithRight += right;
collective += right;
}
max.set(Math.max(max.get(), collective + 1));
int mine = Math.max(mineWithRight, mineWithLeft);
max.set(Math.max(max.get(), mine));
return mine;
}
}
Solution in Python :
class Solution:
def solve(self, root):
res = 0
def consec_path(cur):
if not cur:
return 0, 0
l_inc, l_dec = consec_path(cur.left)
r_inc, r_dec = consec_path(cur.right)
if not cur.left or cur.left.val + 1 != cur.val:
l_inc = 0
if not cur.left or cur.left.val - 1 != cur.val:
l_dec = 0
if not cur.right or cur.right.val + 1 != cur.val:
r_inc = 0
if not cur.right or cur.right.val - 1 != cur.val:
r_dec = 0
nonlocal res
res = max(res, 1 + l_dec + r_inc, 1 + l_inc + r_dec)
return max(1 + r_inc, 1 + l_inc), max(1 + r_dec, 1 + l_dec)
consec_path(root)
return res
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