Complete Binary Tree - Amazon Top Interview Questions


Problem Statement :


Given a binary tree root, return whether it's a complete binary tree.

Constraints

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

Example 1

Input

root = [0, [1, [1, null, null], [0, null, null]], [0, [1, null, null], [0, null, null]]]

Output

True

Example 2

Input

root = [0, [1, [1, null, null], [0, null, null]], [0, null, null]]

Output

True

Example 3

Input

root = [0, [1, null, [0, null, null]], [0, null, null]]

Output

False


Solution :



title-img 




                        Solution in C++ :

bool solve(Tree *root) {
    // okay I read levels and immediately thought of level order traversal ;)
    // but as soon as i reach a point without a child
    // nobody after that should have a child!!!!!!!!

    bool no_kids = false;
    queue<Tree *> q;
    if (root) q.push(root);

    while (not q.empty()) {
        Tree *cur = q.front();
        q.pop();

        if (cur->left)
            if (no_kids)
                return false;
            else
                q.push(cur->left);
        else
            no_kids = true;

        if (cur->right)
            if (no_kids)
                return false;
            else
                q.push(cur->right);
        else
            no_kids = true;
    }

    return true;
}
                    


                        Solution in Java :

import java.util.*;

class Solution {
    public boolean solve(Tree root) {
        return dfs(root, 1) == count;
    }
    int count = 0;
    private int dfs(Tree root, int index) {
        if (root == null)
            return 0;
        count++;
        return Math.max(index, Math.max(dfs(root.left, index * 2), dfs(root.right, index * 2 + 1)));
    }
}
                    


                        Solution in Python : 
                            
class Solution:
    def solve(self, root):
        if root:
            queue = deque([root])
        else:
            return True
        can_child = True

        while queue:
            node = queue.popleft()

            if can_child:
                if node.left:
                    queue.append(node.left)
                else:
                    can_child = False
            else:
                if node.left:
                    return False

            if can_child:
                if node.right:
                    queue.append(node.right)
                else:
                    can_child = False
            else:
                if node.right:
                    return False
        return True


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