Construct Binary Tree from Preorder and Inorder Traversal [LC#105]

Given two integer arrays preorder and inorder where preorder is the preorder traversal of a binary tree and inorder is the inorder traversal of the same tree, construct and return the binary tree.

Intuition

The elements in preorder represents the roots where the inorder traversal needs to be bifurcated. Maintaining a index map of inorder to quickly map to position of any element will allow quickly building the tree recursively.

Code

def build_tree(self, preorder: List[int], inorder: List[int]) -> Optional[TreeNode]:
    val_to_idx = {val: idx for idx, val in enumerate(inorder)}

    def construct(in_left, in_right):
        if in_left > in_right:
            return None

        val = preorder.pop(0)
        in_idx = val_to_idx[val]
        node = TreeNode(val)

        node.left = construct(in_left, in_idx - 1)
        node.right = construct(in_idx + 1, in_right)
        return node
    return construct(0, len(preorder) - 1)

Time Complexity

$T(n) = O(n)$ and $S(n) = O(n)$ for the recursion stack