12. Duplicate subtree in Binary Tree
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The problem can be found at the following link:
To check if a binary tree contains a duplicate subtree of size 2 or more. It does so by checking if there are any identical subtrees within the tree.
Here are the steps in my approach:
The dupSub function does a level-order traversal of the binary tree using a queue to extract all the nodes.
While traversing the tree, it identifies non-leaf nodes and stores them in the nodes vector.
After collecting non-leaf nodes, it iterates through the nodes vector and compares each pair of nodes using the same function to check if they have the same structure and values. If a match is found, it returns true, indicating the presence of a duplicate subtree.
If no duplicate subtree is found during the iteration, it returns false.
Time Complexity: The time complexity of this approach is O(N)
, where N
is the number of nodes in the binary tree. This is because we visit each node once to calculate its height and check the balance condition.
Space Complexity: The space complexity is O(H)
, where H
is the height of the binary tree. In the worst case, the space complexity is O(N + H)
for a skewed tree, and in the best case, it is O(N)
for a balanced tree. This space is used for the recursive call stack.
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