1008. Construct Binary Search Tree from Preorder Traversal
package LeetCode_1008 /** * 1008. Construct Binary Search Tree from Preorder Traversal * https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/description/ * * Return the root node of a binary search tree that matches the given preorder traversal. (Recall that a binary search tree is a binary tree where for every node, any descendant of node. left has a value < node.val, and any descendant of node. right has a value > node.val. Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.) It's guaranteed that for the given test cases there is always possible to find a binary search tree with the given requirements. Example 1: Input: [8,5,1,7,10,12] Output: [8,5,10,1,7,null,12] Constraints: =. 1 <= preorder.length <= 100 =. 1 <= preorder[i] <= 10^8 =. The values of preorder are distinct. * */ class TreeNode(var `val`: Int) { var left: TreeNode? = null var right: TreeNode? = null } class Solution { /* Time complexity: O(nlogn), Space complexity: O(n) * preorder:root->left->right,so the first element of preorder array is the root of tree * */ fun bstFromPreorder(preorder: IntArray): TreeNode? { if (preorder == null || preorder.isEmpty()) { return null } val size = preorder.size val root = TreeNode(preorder[0]) for (i in 1 until size) { help(root, preorder[i]) } return root } private fun help(root: TreeNode?, value: Int): TreeNode? { if (root == null) { return TreeNode(value) } if (root.`val` < value) { root.right = help(root.right, value) } else { root.left = help(root.left, value) } return root } }