Solution

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.

Complexity Analysis

• 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:

!?!../Documents/94_Binary.json:1000,563!?!

Complexity Analysis

• 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

For example:

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].

For more details, please check Threaded binary tree and Explaination of Morris Method

Complexity Analysis

• 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.