Last Updated: 12 Dec, 2020

Left Right Node Balancing

Easy

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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.

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

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Approaches

01 Approach

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:

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

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02 Approach

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). It at any node we encounter that the two values are not equal we return false immediately. If not, we propagate the number of nodes up towards the parent node.

Algorithm:

Create a recursive function isBalanced that takes a node as input and returns the number of nodes in its subtree.
If we reach a null node return 0 as no nodes are present here.
Call the function for left subtree and right subtree recursively.
If the number of nodes in the left and right subtree are unequal we make the result false which is passed to this function by reference.
Else we return the number of nodes as nodes in the left subtree + nodes in the right subtree + 1 (for the current node).
We finally return the value of the result.

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