## Solution

#### Approach 1: Recursion

Intuition

A preorder traversal is:

• (root node) (preorder of left branch) (preorder of right branch)

While a postorder traversal is:

• (postorder of left branch) (postorder of right branch) (root node)

For example, if the final binary tree is [1, 2, 3, 4, 5, 6, 7] (serialized), then the preorder traversal is [1] + [2, 4, 5] + [3, 6, 7], while the postorder traversal is [4, 5, 2] + [6, 7, 3] + [1].

If we knew how many nodes the left branch had, we could partition these arrays as such, and use recursion to generate each branch of the tree.

Algorithm

Let's say the left branch has nodes. We know the head node of that left branch is pre[1], but it also occurs last in the postorder representation of the left branch. So pre[1] = post[L-1] (because of uniqueness of the node values.) Hence, L = post.indexOf(pre[1]) + 1.

Now in our recursion step, the left branch is represnted by pre[1 : L+1] and post[0 : L], while the right branch is represented by pre[L+1 : N] and post[L : N-1].

Complexity Analysis

• Time Complexity: , where is the number of nodes.

• Space Complexity: .

#### Approach 2: Recursion (Space Saving Variant)

Explanation

We present a variation of Approach 1 that uses indexes to refer to the subarrays of pre and post, instead of passing copies of those subarrays. Here, (i0, i1, N) refers to pre[i0:i0+N], post[i1:i1+N].

Complexity Analysis

• Time Complexity: , where is the number of nodes.

• Space Complexity: , the space used by the answer.

Analysis written by: @awice.