Number of ways to divide a Binary tree into two halves

• Last Updated : 09 Dec, 2021

Given a binary tree, the task is to count the number of ways to remove a single edge from the tree such that the tree gets divided into two halves with equal sum.
Examples:

Input:
1
/   \
-1     -1
\
1
Output: 1
Only way to do this will be to remove the edge from the right of the root.
After that we will get 2 sub-trees with sum = 0.
1
/
-1

and

-1
\
1
will be the two sub-trees.

Input:
1
/   \
-1     -1
\
-1
Output: 2

A simple solution will be to remove all the edges of the tree one by one and check if that splits the tree into two halves with the same sum. If it does, we will increase the final answer by 1. This will take O(N2) time in the worst case where “N” is the number of nodes in the tree.
Efficient approach:

1. Create a variable ‘sum’ and store the sum of all the elements of the Binary tree in it. We can find the sum of all the elements of a Binary tree in O(N) time as discussed in this article.
2. Now we perform the following steps recursively starting from root node:
• Find the sum of all the elements of its right sub-tree (“R”). If it’s equal to half of the total sum, we increase the count by 1. This is because removing the edge connecting the current node with its right child will divide the tree into two trees with equal sum.
• Find the sum of all the elements of its left sub-tree (“L”). If it’s equal to half of the total sum, we increase the count by 1.

Below is the implementation of the above approach:

C++

 // C++ implementation of the approach#include using namespace std; // Node of a binary treestruct node {    int data;    node* left;    node* right;    node(int data)    {        this->data = data;        left = NULL;        right = NULL;    }}; // Function to find the sum of// all the nodes of BSTint findSum(node* curr){    // If current node is    // null    if (curr == NULL)        return 0;     // Else    return curr->data + findSum(curr->left)           + findSum(curr->right);} // Function to recursively check// if removing any edge divides tree into// two halvesint checkSum(node* curr, int sum, int& ans){    // Variable to store the    // sum from left and right    // child    int l = 0, r = 0;     // Checking sum from left sub-tree    // if its not null    if (curr->left != NULL) {        l = checkSum(curr->left, sum, ans);        if (2 * l == sum)            ans++;    }     // Checking sum from right sub-tree    // if its not null    if (curr->right != NULL) {        r = checkSum(curr->right, sum, ans);        if (2 * r == sum)            ans++;    }     // Finding the sum of all the elements    // of current node    return l + r + curr->data;} // Function to return the number// of ways to remove an edgeint cntWays(node* root){    // If root is null    if (root == NULL)        return 0;     // To store the final answer    int ans = 0;     // Sum of all the elements of BST    int sum = findSum(root);     // If sum is odd then it won't be possible    // to break it into two halves    if (sum % 2 == 1)        return 0;     // Calling the checkSum function    checkSum(root, sum, ans);     // Returning the final answer    return ans;} // Driver codeint main(){    node* root = new node(1);    root->left = new node(-1);    root->right = new node(-1);    root->right->right = new node(1);     // Print the count of possible ways    cout << cntWays(root);     return 0;}

Java

 // Java implementation of the approachclass GFG{ // Node of a binary treestatic class node{    int data;    node left;    node right;    node(int data)    {        this.data = data;        left = null;        right = null;    }};static int ans; // Function to find the sum of// all the nodes of BSTstatic int findSum(node curr){    // If current node is    // null    if (curr == null)        return 0;     // Else    return curr.data + findSum(curr.left) +                       findSum(curr.right);} // Function to recursively check// if removing any edge divides tree// into two halvesstatic int checkSum(node curr, int sum){    // Variable to store the    // sum from left and right    // child    int l = 0, r = 0;     // Checking sum from left sub-tree    // if its not null    if (curr.left != null)    {        l = checkSum(curr.left, sum);        if (2 * l == sum)            ans++;    }     // Checking sum from right sub-tree    // if its not null    if (curr.right != null)    {        r = checkSum(curr.right, sum);        if (2 * r == sum)            ans++;    }     // Finding the sum of all the elements    // of current node    return l + r + curr.data;} // Function to return the number// of ways to remove an edgestatic int cntWays(node root){    // If root is null    if (root == null)        return 0;     // To store the final answer    ans = 0;     // Sum of all the elements of BST    int sum = findSum(root);     // If sum is odd then it won't be possible    // to break it into two halves    if (sum % 2 == 1)        return 0;     // Calling the checkSum function    checkSum(root, sum);     // Returning the final answer    return ans;} // Driver codepublic static void main(String[] args){    node root = new node(1);    root.left = new node(-1);    root.right = new node(-1);    root.right.right = new node(1);     // Print the count of possible ways    System.out.print(cntWays(root));}} // This code is contributed by PrinciRaj1992

C#

 // C# implementation of the approachusing System; class GFG{ // Node of a binary treepublic class node{    public int data;    public node left;    public node right;    public node(int data)    {        this.data = data;        left = null;        right = null;    }};static int ans; // Function to find the sum of// all the nodes of BSTstatic int findSum(node curr){    // If current node is    // null    if (curr == null)        return 0;     // Else    return curr.data + findSum(curr.left) +                       findSum(curr.right);} // Function to recursively check// if removing any edge divides tree// into two halvesstatic int checkSum(node curr, int sum){    // Variable to store the    // sum from left and right    // child    int l = 0, r = 0;     // Checking sum from left sub-tree    // if its not null    if (curr.left != null)    {        l = checkSum(curr.left, sum);        if (2 * l == sum)            ans++;    }     // Checking sum from right sub-tree    // if its not null    if (curr.right != null)    {        r = checkSum(curr.right, sum);        if (2 * r == sum)            ans++;    }     // Finding the sum of all the elements    // of current node    return l + r + curr.data;} // Function to return the number// of ways to remove an edgestatic int cntWays(node root){    // If root is null    if (root == null)        return 0;     // To store the final answer    ans = 0;     // Sum of all the elements of BST    int sum = findSum(root);     // If sum is odd then it won't be possible    // to break it into two halves    if (sum % 2 == 1)        return 0;     // Calling the checkSum function    checkSum(root, sum);     // Returning the final answer    return ans;} // Driver codepublic static void Main(String[] args){    node root = new node(1);    root.left = new node(-1);    root.right = new node(-1);    root.right.right = new node(1);     // Print the count of possible ways    Console.Write(cntWays(root));}} // This code is contributed by Princi Singh



Python3

 # Python3 implementation of the approach  # Node of a binary treeclass node:    def __init__(self,data):        self.data = data        self.left = None        self.right = None # Function to find the s of# all the nodes of BSTdef findSum(curr):     # If current node is    # null    if (curr == None):        return 0     # Else    return curr.data + findSum(curr.left)+ findSum(curr.right) # Function to recursively check# if removing any edge divides tree into# two halvesdef checkSum(curr, s):    global ans    # Variable to store the    # s from left and right    # child    l = 0; r = 0     # Checking s from left sub-tree    # if its not null    if (curr.left != None):        l = checkSum(curr.left, s)        if (2 * l == s):            ans+=1     # Checking s from right sub-tree    # if its not null    if (curr.right != None):        r = checkSum(curr.right, s)        if (2 * r == s):            ans+=1     # Finding the s of all the elements    # of current node    return l + r + curr.data # Function to return the number# of ways to remove an edgedef cntWays(root):    # If root is null    if (root == None):        return 0     # To store the final answer    global ans    ans = 0     # s of all the elements of BST    s = findSum(root)     # If s is odd then it won't be possible    # to break it into two halves    if (s % 2):        return 0     # Calling the checkSum function    checkSum(root, s)     # Returning the final answer    return ans  # Driver codeif __name__ == '__main__':     root = node(1)    root.left = node(-1)    root.right = node(-1)    root.right.right = node(1)     # Print the count of possible ways    print(cntWays(root))
Output:
1

Time complexity of this approach will be O(N) and space complexity will O(H) where “N” equals number of node in Binary tree and “H” equals height of the Binary Tree.

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