Check if a Given Tree is Mirror Image in K Swaps

Given a binary tree, we can flip (swap the left and right children) the nodes of the tree to obtain a new tree that is a mirror image of the original tree. Our task is to determine if it is possible to make at most K swaps such that the new tree is a mirror image of the original tree.

Approach

We can solve this problem by traversing the tree recursively and swapping the children at each node until we reach a leaf node or the maximum number of swaps is reached.

For each node, we need to check if swapping its children results in a mirror image of the tree. We can do this by traversing both the original and the flipped tree in a pre-order traversal and making sure the node values are the same.

If we reach a leaf node or the maximum number of swaps is reached, we can stop the recursive traversal and return the result.

Algorithm

1. Define a function isMirror(node1, node2) to check if two nodes are a mirror image of each other.
   • Base case: If both nodes are None, return True
   • Recursive case:
     • If either node1 or node2 is None, return False
     • If the values of node1 and node2 are not equal, return False
     • Recursively check if the left child of node1 is a mirror image of the right child of node2 and if the right child of node1 is a mirror image of the left child of node2.
2. Define the function canMirror(root, K) to check if the given tree can be a mirror image in at most K swaps.
   • Base case: If root is None or K is less than 0, return False.
   • If the original tree is a mirror image of the flipped tree, return True.
   • If K is 0, return False.
   • Swap the left and right children of root.
   • Recursively check if the new tree can be a mirror image in at most K-1 swaps.
   • If not, swap the children back to the original positions and repeat the above step with the right child.
   • If neither the left nor the right child can result in a mirror image in at most K-1 swaps, return False.

Code

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def isMirror(self, node1, node2):
        if not node1 and not node2:
            return True
        if not node1 or not node2 or node1.val != node2.val:
            return False
        return self.isMirror(node1.left, node2.right) and self.isMirror(node1.right, node2.left)
        
    def canMirror(self, root: TreeNode, K: int) -> bool:
        if not root or K < 0:
            return False
        if self.isMirror(root, root):
            return True
        if K == 0:
            return False
        root.left, root.right = root.right, root.left
        if self.canMirror(root, K-1):
            return True
        root.left, root.right = root.right, root.left
        return self.canMirror(root.right, K-1)

Complexity Analysis

• Time Complexity: O(N * K), where N is the number of nodes in the tree. For each node, we need to traverse the original and flipped tree in a pre-order traversal, which takes O(N) time. We need to do this at most K times.
• Space Complexity: O(N), where N is the number of nodes in the tree. This is the space used by the recursive stack.