Python program to check if two trees are mirror of each other without using recursion.

Description: The mirror of binary tree is another binary tree with left and right children of all non leaf nodes interchanged.

Example:

Trees in the above figure are mirror of each other.

Program:

class newNode: def __init__(self, data): self.data = data self.left = self.right = None # function to check whether the two binary trees are mirrors of each other or not def areMirrors(root1, root2): st1 = [] st2 = [] while (1): # iterative inorder traversal of 1st tree and reverse inoder traversal of 2nd tree while (root1 and root2): # if the corresponding nodes in the # two traversal have different data # values, then they are not mirrors # of each other. if (root1.data != root2.data): return "No" st1.append(root1) st2.append(root2) root1 = root1.left root2 = root2.right # if at any point one root becomes None and # the other root is not None, then they are # not mirrors. This condition verifies that # structures of tree are mirrors of each other. if (not (root1 == None and root2 == None)): return "No" if (not len(st1) == 0 and not len(st2) == 0): root1 = st1[-1] root2 = st2[-1] st1.pop(-1) st2.pop(-1) # we have visited the node and its left # subtree. Now, it's right subtree's turn root1 = root1.right # we have visited the node and its right # subtree. Now, it's left subtree's turn root2 = root2.left # both the trees have been # completely traversed else: break # tress are mirrors of each other return "Yes" # Driver Code if __name__ == '__main__': # 1st binary tree formation root1 = newNode(1) root1.left = newNode(3) root1.right = newNode(2) root1.right.left = newNode(5) root1.right.right = newNode(4) # 2nd binary tree formation root2 = newNode(1) root2.left = newNode(2) root2.right = newNode(3) root2.left.left = newNode(4) root2.left.right = newNode(5) print(areMirrors(root1, root2))

Output:
Yes
>>>

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