LeetCode in Kotlin

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

/*
 * Example:
 * var ti = TreeNode(5)
 * var v = ti.`val`
 * Definition for a binary tree node.
 * class TreeNode(var `val`: Int) {
 *     var left: TreeNode? = null
 *     var right: TreeNode? = null
 * }
 */
class Solution {
    private var j = 0
    private val map: MutableMap<Int, Int> = HashMap()

    fun get(key: Int): Int {
        return map.getValue(key)
    }

    private fun answer(preorder: IntArray, inorder: IntArray, start: Int, end: Int): TreeNode? {
        if (start > end || j > preorder.size) {
            return null
        }
        val value = preorder[j++]
        val index = get(value)
        val node = TreeNode(value)
        node.left = answer(preorder, inorder, start, index - 1)
        node.right = answer(preorder, inorder, index + 1, end)
        return node
    }

    fun buildTree(preorder: IntArray, inorder: IntArray): TreeNode? {
        j = 0
        for (i in preorder.indices) {
            map[inorder[i]] = i
        }
        return answer(preorder, inorder, 0, preorder.size - 1)
    }
}

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