Left Right Node Balancing

Easy
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Average time to solve is 15m
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Problem statement

You are given a Binary Tree with N nodes. Your task is to check whether every node in the given tree has the same number of nodes on the left and right sides.

For Example : 
Consider the Binary Tree below :

Alt text

The number of nodes on the left side of node 6 is 3 while on the right side it is 1. Hence the answer for the above binary tree is false.
Detailed explanation ( Input/output format, Notes, Images )
Input Format : 
The first line contains a single integer ‘T’ denoting the number of test cases.

The first line of each test case contains the elements of the binary tree in the level order form separated by a single space.

If any node does not have a left or right child, take -1 in its place. Refer to the example below.


1
2 3
-1 4 -1 -1
5 -1
-1 -1

Explanation :
Level 1 :
The root node of the tree is 1

Level 2 :
Left child of 1 = 2
Right child of 1 = 3

Level 3 :
Left child of 2 = null (-1)
Right child of 2 = 4
Left child of 3 = null (-1)
Right child of 3 = null (-1)

Level 4 :
Left child of 4 = 5
Right child of 4 = null (-1)

Level 5 :
Left child of 5 = null (-1)
Right child of 5 = null (-1)

The first not-null node (of the previous level) is treated as the parent of the first two nodes of the current level. The second not-null node (of the previous level) is treated as the parent node for the next two nodes of the current level and so on.

The input ends when all nodes at the last level are null (-1).

Note: The above format was just to provide clarity on how the input is formed for a given tree. 

The sequence will be put together in a single line separated by a single space. Hence, for the above-depicted tree, the input will be given as:

1 2 3 -1 4 -1 -1 5 -1 -1 -1
Output Format : 
For each test case, return True or False denoting whether the given binary tree satisfies the above condition.
Output for each test case will be printed in a separate line.
Note : 
You are not required to print anything; it has already been taken care of. Just implement the function.
Constraints : 
1 <= T <= 100
1 <= number of nodes <= 1000
1 <= nodeVal <= 10^9

Time Limit: 1 sec
Sample Input 1 :
1
2 3 4 -1 -1 -1 -1
Sample Output 1 :
True
Explanation For Sample Output 1 :

Alt text

For the given test case, There is a binary tree of 3 nodes only. Node with value 2 has one node on the left and another on right. Hence each node has the same number of nodes on the left and right parts.
Sample Input 2 :
1
1 2 3 4 5 6 7 -1 -1 -1 -1 -1 -1 -1 -1
Sample Output 2 :
True
Hint

Recursively check for each and every node whether it satisfies the condition or not.

Approaches (2)
Brute Force


 

We can try to find the number of nodes in the subtree of a node as the sum of nodes in the left tree and right subtree + 1 (for the current node). We call another function that calculates the number of nodes in the subtree separately for left and right children.  If at any node we encounter that the two values are not equal we return false immediately. 

 

Algorithm:

 

  1. Create a recursive function leftRightNodeBalancing that takes a node as input and returns whether all nodes satisfy the condition or not.
  2. If we reach a null node return true as no nodes are present here.
  3. Call the function for left subtree and right subtree recursively.
  4. Calculate the number of nodes for left and right sub child using the subtree function.
  5. If the number of nodes in the left and right subtree is unequal we return false.
  6. Else we return the result as logical ‘and’ of results of left and right sub child.

 

Time Complexity

O(N^2), Where N is the number of nodes in the given binary tree.

 

Since we are using a depth-first traversal to visit each and every node of the binary tree. Since we spend O(N) time inside each call of the function to calculate the number of nodes in subtree, we get the overall time complexity as O(N^2).

Space Complexity

O(N), Where N is the number of nodes in the given binary tree.

 

We are performing a depth-first traversal of the tree. So in the worst case (Skewed trees), the maximum recursion depth can reach N. So overall space complexity will be O(N).

Code Solution
(100% EXP penalty)
Left Right Node Balancing
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