323. Construct Binary Search Tree from Preorder Traversal

In a preorder BST stream, each value either becomes the left child of the previous node (if smaller) or the right child of the nearest ancestor it is still larger than. A decreasing stack tracks candidate parents.

Stack of ancestors

Keep a stack of nodes with strictly decreasing values. For each new value v:

• If v < stack.top, it is that node's left child.
• Otherwise pop while stack.top < v; the last popped node is the parent, and v is its right child.

Push v after linking. This builds the BST in one pass.

for v in preorder[1:]:
    if v < stack[-1].val: stack[-1].left = node
    else: while stack and stack[-1].val < v: parent = stack.pop(); parent.right = node
    stack.append(node)

Worked example

[8,5,1,7,10,12] → root 8, left subtree {5,1,7}, right subtree {10,12}.

Complexity

• Time: O(n).
• Space: O(n).

import java.util.ArrayDeque;
import java.util.Deque;

class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        if (preorder.length == 0) return null;
        TreeNode root = new TreeNode(preorder[0]);
        Deque<TreeNode> stack = new ArrayDeque<>();
        stack.push(root);
        for (int i = 1; i < preorder.length; i++) {
            TreeNode node = new TreeNode(preorder[i]);
            if (preorder[i] < stack.peek().val) {
                stack.peek().left = node;
            } else {
                TreeNode parent = stack.peek();
                while (!stack.isEmpty() && stack.peek().val < preorder[i]) {
                    parent = stack.pop();
                }
                parent.right = node;
            }
            stack.push(node);
        }
        return root;
    }
}