Approach 1: Recursive Approach
The first method to solve this problem is using recursion. This is the classical method and is straightforward. We can define a helper function to implement recursion.
Time complexity : . The time complexity is because the recursive function is .
Space complexity : The worst case space required is , and in the average case it's where is number of nodes.
Approach 2: Iterating method using Stack
The strategy is very similiar to the first method, the different is using stack.
Here is an illustration:
Time complexity : .
Space complexity : .
Approach 3: Morris Traversal
In this method, we have to use a new data structure-Threaded Binary Tree, and the strategy is as follows:
Step 1: Initialize current as root
Step 2: While current is not NULL,If current does not have left child a. Add current’s value b. Go to the right, i.e., current = current.right Else a. In current's left subtree, make current the right child of the rightmost node b. Go to this left child, i.e., current = current.left
1 / \ 2 3 / \ / 4 5 6
First, 1 is the root, so initialize 1 as current, 1 has left child which is 2, the current's left subtree is
2 / \ 4 5
So in this subtree, the rightmost node is 5, then make the current(1) as the right child of 5. Set current = cuurent.left (current = 2). The tree now looks like:
2 / \ 4 5 \ 1 \ 3 / 6
For current 2, which has left child 4, we can continue with thesame process as we did above
4 \ 2 \ 5 \ 1 \ 3 / 6
then add 4 because it has no left child, then add 2, 5, 1, 3 one by one, for node 3 which has left child 6, do the same as above. Finally, the inorder taversal is [4,2,5,1,6,3].
Time complexity : . To prove that the time complexity is , the biggest problem lies in finding the time complexity of finding the predecessor nodes of all the nodes in the binary tree. Intuitively, the complexity is , because to find the predecessor node for a single node related to the height of the tree. But in fact, finding the predecessor nodes for all nodes only needs time. Because a binary Tree with nodes has edges, the whole processing for each edges up to 2 times, one is to locate a node, and the other is to find the predecessor node. So the complexity is .
Space complexity : . Arraylist of size is used.