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105. Construct Binary Tree from Preorder and Inorder Traversal

Medium

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.

Example 1:

Input: preorder = [3,9,20,15,7], inorder = [9,3,15,20,7]

Output: [3,9,20,null,null,15,7]

Example 2:

Input: preorder = [-1], inorder = [-1]

Output: [-1]

Constraints:

• 1 <= preorder.length <= 3000
• inorder.length == preorder.length
• -3000 <= preorder[i], inorder[i] <= 3000
• preorder and inorder consist of unique values.
• Each value of inorder also appears in preorder.
• preorder is guaranteed to be the preorder traversal of the tree.
• inorder is guaranteed to be the inorder traversal of the tree.

Solution

import com_github_leetcode.TreeNode

/*
 * Definition for a binary tree node.
 * class TreeNode(_value: Int = 0, _left: TreeNode = null, _right: TreeNode = null) {
 *   var value: Int = _value
 *   var left: TreeNode = _left
 *   var right: TreeNode = _right
 * }
 */
object Solution {
    def buildTree(preorder: Array[Int], inorder: Array[Int]): TreeNode = {
        val map = inorder.zipWithIndex.toMap
        var j = 0

        def answer(start: Int, end: Int): TreeNode = {
            if (start > end || j >= preorder.length) {
                null
            } else {
                val value = preorder(j)
                j += 1
                val index = map(value)
                val node = new TreeNode(value)
                node.left = answer(start, index - 1)
                node.right = answer(index + 1, end)
                node
            }
        }

        answer(0, preorder.length - 1)
    }
}

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