Check if removing an edge can divide a Binary Tree in two halves - GeeksforGeeks
Given a Binary Tree, find if there exist edge whose removal creates two trees of equal size.
O(n2)
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Given a Binary Tree, find if there exist edge whose removal creates two trees of equal size.
Input : root of following tree
5
/ \
1 6
/ / \
3 7 4
Output : true
Removing edge 5-6 creates two trees of equal size
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.
// 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 treeint 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;}O(n2)
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.
// To calculate size of tree with given rootint 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 treebool 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);}
