1008. Construct Binary Search Tree from Preorder Traversal

#### QUESTION:

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]

Note:

1 <= preorder.length <= 100 The values of preorder are distinct.

#### SOLUTION:

``````class Solution {
public TreeNode bstFromPreorder(int[] preorder) {
if(preorder.length==0)return null;
TreeNode root = bstFromPreorderHelper(preorder,0,preorder.length-1);
return root;
}
public static TreeNode bstFromPreorderHelper(int[] preorder,int start,int end){
if(start==end) return new TreeNode(preorder[start]);
if(start>end) return null;
TreeNode root = new TreeNode(preorder[start]);
int index = -1;
for(int i = start+1;i<=end;i++){
if(preorder[i]>preorder[start]){
index = i;
break;
}
}
if(index == -1)
root.left = bstFromPreorderHelper(preorder,start+1,end);
else if(index==start+1)
root.right = bstFromPreorderHelper(preorder,index,end);
else{
root.left = bstFromPreorderHelper(preorder,start+1,index-1);
root.right = bstFromPreorderHelper(preorder,index,end);
}

return root;
}
}
``````
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