1008. Construct Binary Search Tree from Preorder Traversal

Description:

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.

 

 

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode bstFromPreorder(int[] preorder) {
        TreeNode root=new TreeNode(preorder[0],null,null);
        for(int i=1;i<preorder.length;++i){
            insert(preorder[i],root);
        }
        return root;
    }
    
    public void insert(int val, TreeNode root){
        if(root==null)
            return;
        
        if(val<root.val){
            if(root.left==null)
                root.left=new TreeNode(val);
            else
                insert(val,root.left);
        }
        else if(val>root.val){
            if(root.right==null)
                root.right=new TreeNode(val);
            else
                insert(val,root.right);
        }
    }
}

// when n==preorder.length,
// T/C : O(nlogn)
// S/C: O(n)