Construct BST from given preorder traversal

Construct BST from given preorder traversal | Set 1 | GeeksforGeeks
Method 2 ( O(n) time complexity )

The idea used here is inspired from method 3 of this post. The trick is to set a range {min .. max} for every node. Initialize the range as {INT_MIN .. INT_MAX}. The first node will definitely be in range, so create root node. To construct the left subtree, set the range as {INT_MIN …root->data}. If a values is in the range {INT_MIN .. root->data}, the values is part part of left subtree. To construct the right subtree, set the range as {root->data..max .. INT_MAX}.

class Index {

 

    int index = 0;

}

 

    Index index = new Index();

 

    // A recursive function to construct BST from pre[]. preIndex is used

    // to keep track of index in pre[].

    Node constructTreeUtil(int pre[], Index preIndex, int key,

            int min, int max, int size) {

 

        // Base case

        if (preIndex.index >= size) {

            return null;

        }

 

        Node root = null;

 

        // If current element of pre[] is in range, then

        // only it is part of current subtree

        if (key > min && key < max) {

 

            // Allocate memory for root of this subtree and increment *preIndex

            root = new Node(key);

            preIndex.index = preIndex.index + 1;

 

            if (preIndex.index < size) {

 

                // Contruct the subtree under root

                // All nodes which are in range {min .. key} will go in left

                // subtree, and first such node will be root of left subtree.

                root.left = constructTreeUtil(pre, preIndex, pre[preIndex.index],

                        min, key, size);

 

                // All nodes which are in range {key..max} will go in right

                // subtree, and first such node will be root of right subtree.

                root.right = constructTreeUtil(pre, preIndex, pre[preIndex.index],

                        key, max, size);

            }

        }

 

        return root;

    }

 

    // The main function to construct BST from given preorder traversal.

    // This function mainly uses constructTreeUtil()

    Node constructTree(int pre[], int size) {

        int preIndex = 0;

        return constructTreeUtil(pre, index, pre[0], Integer.MIN_VALUE,

                Integer.MAX_VALUE, size);

    }

http://algorithms.tutorialhorizon.com/construct-binary-search-tree-from-a-given-preorder-traversal-using-recursion/

Solu­tion to the prob­lem is sim­i­lar to isBST Max-Min Solu­tion.

“Your root value can have any value between -∞ to + ∞, say it is 30 here, When you val­i­date the right child of 30, it can take any value between 30 and + ∞. When you val­i­date the left child of 30, it can take any value between — ∞ and 30. likewise

So the idea is Pass the min­i­mum and max­i­mum val­ues between which the node’s value must lie.

• Exam­ple: int[] pre­Order = { 20, 10, 5, 1, 7, 15, 30, 25, 35, 32, 40 };
• First ele­ment in the pre­order[] will def­i­nitely be the root, which is 20 here.
• we start with the range min­i­mum = Integer.MIN_VALUE and max­i­mum = Interger.MAX_VALUE, so your root can take any value between this range.
• So when putting the left node of 20(root), it must lie within the range to min­i­mum = Integer.MIN_VALUE and max­i­mum = 20, so check the next ele­ment in the pre­order[], if it lies in this range, make it the left child to the root, else it must the the right chlid of the root and so on. See the fig­ure for bet­ter under­stand­ing. ( see the exe­cu­tion sequence)

47 public int pos = 0;
48 public static TreeNode preorderToBST(int[] preorder, int min, int max){
49     if(preorder == null || preorder.length == 0 || pos == preorder.length) return null;
50     int val = preorder[pos];
51     if(min < val && val < max){
52         TreeNode root = new TreeNode(val);
53         pos ++;
54         root.left = preorderToBST(preorder, min, val);
55         root.right = preorderToBST(preorder, val, max);
56     }
57 }

X. Stack
http://www.cnblogs.com/reynold-lei/p/4367903.html

 2 class BSTNode{
 3     public TreeNode tree;
 4     public Integer min;
 5     public Integer max;
 6     public BSTNode(TreeNode tree, int min, int max){
 7         this.tree = tree;
 8         this.min = min;
 9         this.max = max;
10     }
11 }
15 public static TreeNode preorderToBST(int[] preorder){
16     if(preorder == null || preorder.length == 0) return null;
17     LinkedList<BSTNode> stack = new LinkedList<BSTNode>();
18     int len = preorder.length;
19     TreeNode result = new TreeNode(preorder[0]);
20     BSTNode root = new BSTNode(result, Integer.MIN_VALUE, Integer.MAX_VALUE);
21     stack.push(root);
22     for(int i = 1; i < len; i ++){
23         TreeNode cur = new TreeNode(preorder[i]);
24         while(!stack.isEmpty()){
25             BSTNode tmp = stack.peek();
26             if(tmp.min < cur.val && cur.val < tmp.tree.val){//left child
27                 tmp.tree.left = cur;
28                 stack.push(new BSTNode(cur, tmp.min, tmp.tree.val));
29                 break;
30             }else if(tmp.tree.val < cur.val && cur.val < tmp.max){//right child
31                 tmp.tree.right = cur;
32                 stack.push(new BSTNode(cur, tmp.tree.val, tmp.max));
33                 break;
34             }else if(cur.val > tmp.max){//not this treenode's child
35                 stack.pop();
36             }else{
37                 System.out.println("Error happens! This is not a valid preorder traersal array. ");
38                 return null;
39             }
40         }
41     }
42     return result;
43 }

http://www.geeksforgeeks.org/construct-bst-from-given-preorder-traversal-set-2/

    Node constructTree(int pre[], int size) {

 

        // The first element of pre[] is always root

        Node root = new Node(pre[0]);

 

        Stack<Node> s = new Stack<Node>();

 

        // Push root

        s.push(root);

 

        // Iterate through rest of the size-1 items of given preorder array

        for (int i = 1; i < size; ++i) {

            Node temp = null;

 

            /* Keep on popping while the next value is greater than

             stack's top value. */

            while (!s.isEmpty() && pre[i] > s.peek().data) {

                temp = s.pop();

            }

 

            // Make this greater value as the right child and push it to the stack

            if (temp != null) {

                temp.right = new Node(pre[i]);

                s.push(temp.right);

            }

             

            // If the next value is less than the stack's top value, make this value

            // as the left child of the stack's top node. Push the new node to stack

            else {

                temp = s.peek();

                temp.left = new Node(pre[i]);

                s.push(temp.left);

            }

        }

 

        return root;

    }

http://algorithms.tutorialhorizon.com/construct-binary-search-tree-from-a-given-preorder-traversal-using-recursion/
http://www.fusu.us/2013/07/re-creating-binary-search-tree-given.html

private static int index;

private static Node binaryPreorderToTree(


int[] arr, int length, int min, int max) {




if (index >= length) {


return null;


}




Node root = null;




int currentNode = arr[index];




if (currentNode > min && currentNode < max) {


root = new Node(currentNode);


index++;




if (index < length) {


root.left = 


binaryPreorderToTree(arr, length, min, currentNode);


}




if (index < length) {


root.right = 


binaryPreorderToTree(arr, length, currentNode, max);


}


}




return root;


}


public static Node binaryPreorderToTree(int[] arr, int length) {


if (length <= 0) {


return null;


}




return binaryPreorderToTree(


arr, length, Integer.MIN_VALUE, Integer.MAX_VALUE);


}

Method 1 ( O(n^2) time complexity )
The first element of preorder traversal is always root. We first construct the root. Then we find the index of first element which is greater than root. Let the index be ‘i’. The values between root and ‘i’ will be part of left subtree, and the values between ‘i+1′ and ‘n-1′ will be part of right subtree. Divide given pre[] at index “i” and recur for left and right sub-trees.

struct node* constructTreeUtil (int pre[], int* preIndex,

                                int low, int high, int size)

{

    // Base case

    if (*preIndex >= size || low > high)

        return NULL;

    // The first node in preorder traversal is root. So take the node at

    // preIndex from pre[] and make it root, and increment preIndex

    struct node* root = newNode ( pre[*preIndex] );

    *preIndex = *preIndex + 1;

    // If the current subarry has only one element, no need to recur

    if (low == high)

        return root;

    // Search for the first element greater than root

    int i;

    for ( i = low; i <= high; ++i )

        if ( pre[ i ] > root->data )

            break;

    // Use the index of element found in postorder to divide postorder array in

    // two parts. Left subtree and right subtree

    root->left = constructTreeUtil ( pre, preIndex, *preIndex, i - 1, size );

    root->right = constructTreeUtil ( pre, preIndex, i, high, size );

    return root;

}

// The main function to construct BST from given preorder traversal.

// This function mainly uses constructTreeUtil()

struct node *constructTree (int pre[], int size)

{

    int preIndex = 0;

    return constructTreeUtil (pre, &preIndex, 0, size - 1, size);

}

Read full article from Construct BST from given preorder traversal | Set 1 | GeeksforGeeks