# Check if removing an edge can divide a Binary Tree in two halves

Given a Binary Tree, find if there exist edge whose removal creates two trees of equal size.

Examples:

`Input : root of following tree            5          /   /        1      6           /      /  /      3      7    4 Output : true Removing edge 5-6 creates two trees of equal size   Input : root of following tree            5          /   /        1      6                 /  /            7    4          /  /    /         3    2    8 Output : false There is no edge whose removal creates two trees of equal size.`

Source- Kshitij IIT KGP

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Method 1 (Simple)

First count number of nodes in whole tree. Let count of all nodes be n. Now traverse tree and for every node, find size of subtree rooted with this node. Let the subtree size be s. If n-s is equal to s, then return true, else false.

`// C++ program to check if there exist an edge whose // removal creates two trees of same size #include<bits/stdc++.h> using namespace std;  struct Node {     int data;     struct Node* left, *right; };  // utility function to create a new node struct Node* newNode(int x) {     struct Node* temp = new Node;     temp->data = x;     temp->left = temp->right = NULL;     return temp; };  // To calculate size of tree with given root int count(Node* root) {     if (root==NULL)         return 0;     return count(root->left) + count(root->right) + 1; }  // This function returns true if there is an edge // whose removal can divide the tree in two halves // n is size of tree bool checkRec(Node* root, int n) {     // Base cases     if (root ==NULL)        return false;      // Check for root     if (count(root) == n-count(root))         return true;      // Check for rest of the nodes     return checkRec(root->left, n) ||            checkRec(root->right, n); }  // This function mainly uses checkRec() bool check(Node *root) {     // Count total nodes in given tree     int n = count(root);      // Now recursively check all nodes     return checkRec(root, n); }  // Driver code int main() {     struct Node* root = newNode(5);     root->left = newNode(1);     root->right = newNode(6);     root->left->left = newNode(3);     root->right->left = newNode(7);     root->right->right = newNode(4);      check(root)?  printf("YES") : printf("NO");      return 0; }`

Output :

`YES`

Time complexity of above solution is O(n 2 ) where n is number of nodes in given Binary Tree.

Method 2 (Efficient)

We can find the solution in O(n) time. The idea is to traverse tree in bottom up manner and while traversing keep updating size and keep checking if there is a node that follows the required property.

Below is C++ implementation of above idea.

`// C++ program to check if there exist an edge whose // removal creates two trees of same size #include<bits/stdc++.h> using namespace std;  struct Node {     int data;     struct Node* left, *right; };  // utility function to create a new node struct Node* newNode(int x) {     struct Node* temp = new Node;     temp->data = x;     temp->left = temp->right = NULL;     return temp; };  // To calculate size of tree with given root int count(Node* root) {     if (root==NULL)         return 0;     return count(root->left) + count(root->right) + 1; }  // This function returns size of tree rooted with given // root. It also set "res" as true if there is an edge // whose removal divides tree in two halves. // n is size of tree int checkRec(Node* root, int n, bool &res) {     // Base case     if (root == NULL)        return 0;      // Compute sizes of left and right children     int c = checkRec(root->left, n, res) + 1 +             checkRec(root->right, n, res);      // If required property is true for current node     // set "res" as true     if (c == n-c)         res = true;      // Return size     return c; }  // This function mainly uses checkRec() bool check(Node *root) {     // Count total nodes in given tree     int n = count(root);      // Initialize result and recursively check all nodes     bool res = false;     checkRec(root, n,  res);      return res; }  // Driver code int main() {     struct Node* root = newNode(5);     root->left = newNode(1);     root->right = newNode(6);     root->left->left = newNode(3);     root->right->left = newNode(7);     root->right->right = newNode(4);      check(root)?  printf("YES") : printf("NO");      return 0; }`

Output :

`YES`