1008. Construct Binary Search Tree from Preorder Traversal

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Problem

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]

Code

class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        return helper(preorder, 0, preorder.length - 1);
    }

    private TreeNode helper(int[] preorder, int start, int end) {
        if (start > end) return null;
        TreeNode root = new TreeNode(preorder[start]);
        // first larger number
        for (int i = start + 1; i <= end; i++) {
            int num = preorder[i];
            if (num > preorder[start]) {
                root.left = helper(preorder, start + 1, i - 1);
                root.right = helper(preorder, i, end);
                return root;
            }
        }

        // no first larger number
        root.left = helper(preorder, start + 1, end);
        root.right = null;

        return root;
    }
}