Friday, November 21, 2014

Foldable Binary Trees - Given a binary tree, find out if the tree can be folded or not.

Problem

Given a binary tree, find out if the tree can be folded or not.
A tree can be folded if left and right subtrees of the tree are structure wise mirror image of each other. An empty tree is considered as foldable.

Examples


Consider the below trees:
(a) and (b) can be folded.
(c) and (d) cannot be folded.

(a)
       10
     /    \
    7      15
     \    /
      9  11

(b)
        10
       /  \
      7    15
     /      \
    9       11

(c)
        10
       /  \
      7   15
     /    /
    5   11

(d)

         10
       /   \
      7     15
    /  \    /
   9   10  12

Solution


Method 1 (Change Left subtree to its Mirror and compare it with Right subtree)
Algorithm: isFoldable(root)
1) If tree is empty, then return true.
2) Convert the left subtree to its mirror image
    mirror(root->left); /* See this post */
3) Check if the structure of left subtree and right subtree is same
   and store the result.
    res = isStructSame(root->left, root->right); /*isStructSame()
        recursively compares structures of two subtrees and returns
        true if structures are same */
4) Revert the changes made in step (2) to get the original tree.
    mirror(root->left);
5) Return result res stored in step 2. 


Method 2 - Using recursion
here is the pseudocode:

boolean mirrorEquals(BTree left, BTree right) {
  if (left == null || right == null) 
             return left == null && right == null;
  return left.value == right.value && mirrorEquals(left.left, right.right) && mirrorEquals(left.right, right.left);
}

Method 3 - Iteratively

bool isMirrorItselfIteratively(BinaryTreeNode *root) 
{
    /// use single queue
    if(!root) return true;
    queue<BinaryTreeNode> q;
    q.push(root.left);
    q.push(root.right);
    BinaryTreeNode l, r;
    while(!q.empty()) {
        l = q.front();
        q.pop();
        r = q.front();
        q.pop();
        if(l==NULL && r==NULL) continue;
        if(l==NULL || r==NULL || l.data!=r.data) return false;
        q.push(l.left);
        q.push(r.right);
        q.push(l.right);
        q.push(r.left);
    }

    return true;
}


Thanks.

References 

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