Monday, March 28, 2016

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


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.
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 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;
}

 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 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);
}

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