1008. Construct Binary Search Tree from Preorder Traversal

Medium

Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.

It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.

A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.

A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.

Example 1:

Input: preorder = [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

Example 2:

Input: preorder = [1,3]
Output: [1,null,3]

Constraints:

• 1 <= preorder.length <= 100

• 1 <= preorder[i] <= 1000

• All the values of preorder are unique.

解題

Runtime: 0 ms, faster than 100%

Memory Usage: 2.3 MB, less than 100%

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func bstFromPreorder(preorder []int) *TreeNode {
    // preorder -> 中 左 右
    if len(preorder) == 0 {
        return nil
    }

    i := 1
    for i=1; i<len(preorder); i++ {
        if preorder[i] >  preorder[0] {
            break
        }
    }

    return &TreeNode{ Val: preorder[0], Left: bstFromPreorder(preorder[1:i]), Right: bstFromPreorder(preorder[i:]) }
}