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May Leetcode Challenge - Day 24

Construct Binary Search Tree from Preorder Traversal

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]

1266

Constraints:

  • 1 <= preorder.length <= 100
  • 1 <= preorder[i] <= 10^8
  • The values of preorder are distinct.

Solution

Do in-order traversal with range limit
Range starts with (Int.min, Int.max)

class Solution {
    func bstFromPreorder(_ preorder: [Int]) -> TreeNode? {
        var data = preorder
        var index = 0
        return helper(&data, Int.min, Int.max, &index)
    }
    
    func helper(_ data: inout [Int], _ min: Int, _ max: Int, _ cIndex: inout Int) -> TreeNode? {
        guard cIndex < data.count else {
            return nil
        }
        
        guard data[cIndex] > min && data[cIndex] < max else {
            return nil
        }
        
        let root = TreeNode(data[cIndex])
        cIndex += 1
        root.left = helper(&data, min, data[cIndex - 1], &cIndex)
        root.right = helper(&data, data[cIndex - 1], max, &cIndex)

        return root
    }
}

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