Vertical Lines in Binary Tree - Amazon Top Interview Questions


Problem Statement :


Given a binary tree root, return the number of unique vertical lines that can be drawn such that every node has a line intersecting it. Each left child is angled at 45 degrees to its left, while the right child is angled at 45 degrees to the right.

For example, root and root.left.right are on the same vertical line.

Constraints

1 ≤ n ≤ 100,000 where n is the number of nodes in root

Example 1

Input

root = [1, [2, [3, null, null], null], [4, null, [5, null, null]]]

Output

5

Explanation

There's a unique vertical line over every node.

Example 2

Input

root = [1, null, [2, null, [3, null, [4, null, null]]]]

Output

4


Solution :



title-img 




                        Solution in C++ :

unordered_set<int> s;
void dfs(Tree* t, int d) {
    if (t == NULL) return;
    s.insert(d);
    dfs(t->left, d - 1);
    dfs(t->right, d + 1);
}
int solve(Tree* root) {
    s.clear();
    dfs(root, 0);
    return s.size();
}
                    


                        Solution in Java :

import java.util.*;

/**
 * public class Tree {
 *   int val;
 *   Tree left;
 *   Tree right;
 * }
 */
class Solution {
    static int L;
    static int R;
    public int solve(Tree root) {
        L = 0;
        R = 0;
        helper(root, 0);
        return (R - L + 1);
    }

    public void helper(Tree t, int p) {
        L = Math.min(L, p);
        R = Math.max(R, p);
        if (t.left != null)
            helper(t.left, p - 1);
        if (t.right != null)
            helper(t.right, p + 1);
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, root):
        left = right = 0

        def dfs(root, x):
            nonlocal left, right
            if not root:
                return
            left = min(x, left)
            right = max(x, right)
            dfs(root.left, x - 1)
            dfs(root.right, x + 1)

        dfs(root, 0)
        return right - left + 1


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