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# construct-binary-search-tree-from-preorder-traversal

## Problem 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.)
Example 1:
Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]
1266
Note:
1 <= preorder.length <= 100
The values of preorder are distinct.

## Solution

Binary Search Tree, `left.val < root.val < right.val` preorder (root -> left -> right) means the first is root value, recursively traverse array:
• if smaller than root value, left tree
• if bigger than root value, right tree
For example:
Construct Binary Search Tree from Preorder Traversal

### Complexity Analysis

Time Complexity: `O(N)`
• N - the length of array nums

### Code

class Solution {
public TreeNode bstFromPreorder(int[] preorder) {
return buildBST(preorder, Integer.MAX_VALUE);
}
int idx = 0;
private TreeNode buildBST(int[] preorder, int limit) {
if (idx == preorder.length || preorder[idx] > limit) return null;
TreeNode root = new TreeNode(preorder[idx++]);
// element value < root.val, add to root.left, limit set to root.val
root.left = buildBST(preorder, root.val);
root.right = buildBST(preorder, limit);
return root;
}
}