Given a binary tree, return all duplicate subtrees. For each kind of duplicate subtrees, you only need to return the root node of any one of them.
Two trees are duplicate if they have the same structure with same node values.
1 / \ 2 3 / / \ 4 2 4 / 4The following are two duplicate subtrees:
2 / 4and
4Therefore, you need to return above trees' root in the form of a list.
We can serialize each subtree. For example, the tree\n
1\n / \\\n 2 3\n / \\\n 4 5\n
can be represented as the serialization
1,2,#,#,3,4,#,#,5,#,#, which is a unique representation of the tree.
Perform a depth-first search, where the recursive function returns the serialization of the tree. At each node, record the result in a map, and analyze the map after to determine duplicate subtrees.\n\n
Time Complexity: , where is the number of nodes in the tree. We visit each node once, but each creation of
serial may take work.
Space Complexity: , the size of
Suppose we have a unique identifier for subtrees: two subtrees are the same if and only if they have the same id.\n
Then, for a node with left child id of
x and right child id of
(node.val, x, y) uniquely determines the tree.
If we have seen the triple
(node.val, x, y) before, we can use the identifier we\'ve remembered. Otherwise, we\'ll create a new one.
Time Complexity: , where is the number of nodes in the tree. We visit each node once.\n
Space Complexity: . Every structure we use is using storage per node.\n