Construct BST from given preorder traversal | Set 1

Given preorder traversal of a binary search tree, construct the BST.

For example, if the given traversal is {10, 5, 1, 7, 40, 50}, then the output should be root of following tree.

     10
   /   \
  5     40
 /  \      \
1    7      50

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.
For example in {10, 5, 1, 7, 40, 50}, 10 is the first element, so we make it root. Now we look for the first element greater than 10, we find 40. So we know the structure of BST is as following.

             10
           /    \
          /      \
  {5, 1, 7}       {40, 50}

We recursively follow above steps for subarrays {5, 1, 7} and {40, 50}, and get the complete tree.

/* A O(n^2) program for construction of BST from preorder traversal */
#include <stdio.h>
#include <stdlib.h>

/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node
{
    int data;
    struct node *left;
    struct node *right;
};

// A utility function to create a node
struct node* newNode (int data)
{
    struct node* temp = (struct node *) malloc( sizeof(struct node) );

    temp->data = data;
    temp->left = temp->right = NULL;

    return temp;
}

// A recursive function to construct Full from pre[]. preIndex is used
// to keep track of index in pre[].
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 preorder to divide preorder 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);
}

// A utility function to print inorder traversal of a Binary Tree
void printInorder (struct node* node)
{
    if (node == NULL)
        return;
    printInorder(node->left);
    printf("%d ", node->data);
    printInorder(node->right);
}

// Driver program to test above functions
int main ()
{
    int pre[] = {10, 5, 1, 7, 40, 50};
    int size = sizeof( pre ) / sizeof( pre[0] );

    struct node *root = constructTree(pre, size);

    printf("Inorder traversal of the constructed tree: \n");
    printInorder(root);

    return 0;
}

Output:

 1 5 7 10 40 50

Time Complexity: O(n^2)



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}.

/* A O(n) program for construction of BST from preorder traversal */
#include <stdio.h>
#include <stdlib.h>
#include <limits.h>

/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node
{
    int data;
    struct node *left;
    struct node *right;
};

// A utility function to create a node
struct node* newNode (int data)
{
    struct node* temp = (struct node *) malloc( sizeof(struct node) );

    temp->data = data;
    temp->left = temp->right = NULL;

    return temp;
}

// A recursive function to construct BST from pre[]. preIndex is used
// to keep track of index in pre[].
struct node* constructTreeUtil( int pre[], int* preIndex, int key,
                                int min, int max, int size )
{
    // Base case
    if( *preIndex >= size )
        return NULL;
 
    struct 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 = newNode ( key );
        *preIndex = *preIndex + 1;
        
        if (*preIndex < 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],
                                        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],
                                         key, max, 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, pre[0], INT_MIN, INT_MAX, size );
}

// A utility function to print inorder traversal of a Binary Tree
void printInorder (struct node* node)
{
    if (node == NULL)
        return;
    printInorder(node->left);
    printf("%d ", node->data);
    printInorder(node->right);
}

// Driver program to test above functions
int main ()
{
    int pre[] = {10, 5, 1, 7, 40, 50};
    int size = sizeof( pre ) / sizeof( pre[0] );

    struct node *root = constructTree(pre, size);

    printf("Inorder traversal of the constructed tree: \n");
    printInorder(root);

    return 0;
}

Output:

 1 5 7 10 40 50

Time Complexity: O(n)

We will soon publish a O(n) iterative solution as a separate post.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.





  • prashant jha

    the o(n^2) recusive solution is here

    #include
    #define size 50
    using namespace std;
    struct tnode
    {
    tnode* lchild;
    int data;
    tnode* rchild;
    tnode(int d)
    {
    lchild=NULL;
    data=d;
    rchild=NULL;
    }
    };
    int search_rci(int n,int pre[],int low,int high)
    {
    for(int i=low;in)
    return i;
    }
    return -1;
    }
    int search_lci(int n,int pre[],int low,int high)
    {
    for(int i=low;i<=high;i++)
    {
    if(pre[i]<n)
    return i;
    }
    return -1;
    }
    tnode* fun(tnode* root,int pre[],int low,int high)
    {
    if((low<0)||(highlchild=fun(root->lchild,pre,lci,rci-1);
    root->rchild=fun(root->rchild,pre,rci,high);
    return root;
    }

    void display(tnode* root)
    {
    if(root)
    {
    display(root->lchild);
    cout<data<rchild);
    }
    }
    int main()
    {
    tnode* root=NULL;
    int low,high;
    int pre[]= {10, 5, 1, 7, 40, 50};
    int n=sizeof(pre)/sizeof(pre[0]);
    display(fun(root,pre,0,n-1));
    return 0;
    }

  • danny

    We can also have another O(nlogn) Time Complexity solution but it uses O(n) Space Complexity also…

    Step-1: Sort the preorder[] array and store it in another inorder[] array.
    Step-2: Construct the BST using preorder[] and inorder[] array.

    O(n) Space Complexity….
    Please correct me if I am wrong

  • Guest

    We can also have another O(n) Time Complexity solution but it uses O(n) Space Complexity also…

    Step-1: Sort the preorder[] array and store it in another inorder[] array.
    Step-2: Construct the BST using preorder[] and inorder[] array.

    O(n) Space and Time Complexity….
    Please correct me if I am wrong

  • Rg

    Method 2: Why do you need min? Since this is a preorder traversal only max is enough. Can you give me a counter example where you would get in incorrect BST if you don’t pass min around

  • Amit Bgl

    wow code 😀

  • Shubham Singhal

    Can we just traverse the input array, and insert nodes in BST in that order and get required BST.

     
    /* Paste your code here (You may delete these lines if not writing code) */
     
    • Roman Dzhabarov

      I think it will work but having O(n*n) in the worst case and O(n*log(n)) in average. Suppose, 9,8,7,6,5,4 etc preorder.

  • abhishek08aug

    Intelligent 😀

  • Soumya Sengupta

    When we execute the statement…
    root->right = constructTreeUtil ( pre, preIndex, i, high, size );
    preIndex should point to i…..so dat i becomes the root and we proceed further……
    isnt it?

  • code_ignitor

    Should’nt we include these lines..

     
    if(min>max || key<min || key > max)
         return NULL;
     
    • code_ignitor

      Sorry !! Got your code :)

  • Balasubramanian

    Why is it that we are passing the parameter-“key” to the function. Couldn’t we just pass preIndex as a reference and increment it when we create a new node.
    That way, we can just use pre[preIndex] to create an element.

    Sample code:

     
    node *buildBST(int pre[],int &idx,int size,int minVal,int maxVal){
    	if(idx==size)
    		return NULL;
    	node *root=NULL;
    	if(minVal<pre[idx] && pre[idx]<maxVal){
    		root=new node(pre[idx++]); // incrementing idx
    		root->left=buildBST(pre,idx,size,minVal,root->data);
    		root->right=buildBST(pre,idx,size,root->data,maxVal);
    	}
    	return root;
    }
     

    Please correct me if I am wrong.
    Thanks,

  • prakash

    hi geeks,
    when preindex reaches “size”, you will get segmentation fault (pre[*preindex]) in
    the following call::
    root->left = constructTreeUtil( pre, preIndex, pre[*preIndex],
    min, key, size );

    • http://www.facebook.com/barnwal.aashish Aashish

      Please take a closer look at the code. When the value of *preIndex = size, the function returns NULL.

       
      // Base case
          if (*preIndex >= size || low > high)
              return NULL;
       
      • prakash

        fine. but while passing argument,ie passing key value in argument, pre[*preindex] is out of bound access when *preindex==size, since pre[size -1] is the last element in array.

        • http://www.facebook.com/barnwal.aashish Aashish

          Thanks for pointing this out. We have updated the code. Keep it up.

  • Nomind

    #include
    #include
    #include

    /* A binary tree node has key, left child and right child */
    struct node
    {
    int key;
    struct node* left;
    struct node* right;
    };

    /* Helper function that allocates a new node with the given key and
    NULL left and right pointers.*/
    struct node* newNode(int key)
    {
    struct node* node = (struct node*)malloc(sizeof(struct node));
    node->key = key;
    node->left = NULL;
    node->right = NULL;
    return(node);
    }

    // Function to find ceil of a given input in BST. If input is more
    // than the max key in BST, return -1
    int floor(node *root, int input)
    {
    // Base case
    if( root == NULL )
    return INT_MAX;

    // We found equal key
    if( root->key == input )
    return root->key;

    // If root’s key is smaller, ceil must be in right subtree
    if( root->key > input )
    return floor(root->left, input);

    // Else, either left subtree or root has the ceil value
    int Floor = floor(root->right, input);
    return (Floor key;
    }

    // Driver program to test above function
    int main()
    {
    node *root = newNode(8);

    root->left = newNode(4);
    root->right = newNode(12);

    root->left->left = newNode(2);
    root->left->right = newNode(6);

    root->right->left = newNode(10);
    root->right->right = newNode(14);

    for(int i = 0; i < 16; i++)
    printf("%d %d\n", i, floor(root, i));
    getchar();
    return 0;
    }

  • Rakesh
     
    I have doubt regarding the problem.
    Pre Order traversal of BST will not give unique tree.
    taking above example.
    
    {10, 5, 1, 7, 40, 50}
    can be pre order traversal of any of below trees
           10                               10
         5    40                         5      40
       1         50       OR          1     7 
         7
    
    i.e 7 can be right child of 1 or 5
    
    Can someone please explain the problem?
    Is it ok to just create any BST from the array?
     
    • Aashish

      The problem lies in the first BST. Observe node 7. It should lie on the right side of node 5 only.

  • Animesh Pratap Singh
     
    
    
    
    method 1 does not compute in O(n^2) time, as it is a recursive function and may split at-most into 2 parts.
    
    1- root will split into at-most ( binary tree)
    2- suppose the 1st largest no. (larger than root) after root 
       occurs exactly in middle
    
       eg- 40,1,2,4,(30),5,6,7,8 
    
    3- then
         T(n)= 2T(n/2)+ O(n)
    
          {  f(n)=O(n) suppose in worst case (30) ie largest 
             element than root occurs at last
          } 
    
    4= by master's theorem
    
        time complexity = O(n logn)....
    
    
     
    • Aashish

      The method#1 will run in NlogN time in best case( when the index to be splitted happens to be in the mid-way always ).
      Think about the below tree( Skew Tree ):

      10, 8, 7, 6

      Observe that in case of skew tree, it can split one element on one side and the rest on other side

      For each index, the for loop runs for O(N).
      The time complexity can be given by:
      N-1 + N-2 + ….. + 1
      Which is asymptotically equivalent to O(N^2).

  • Sairam

    I think there is no need to use preIndex variable in Method 1. Instead of passing *preindex, just pass (low + 1) in the recursive function. Correct me if I am wrong.

     
    /* Paste your code here (You may delete these lines if not writing code) */
     
  • hard2code

    I hope the following code would work in O(n). Is it not?

    public Node buildTree(int[] arr, int l, int r) {
    if(r-l < 0 || l = arr.length)
    return null;
    Node node = new Node(arr[l]);
    int lr =l;
    while(lr < r && arr[lr+1] < arr[l])
    lr++;
    node.left = buildTree(arr, l+1, lr);
    node.right = buildTree(arr, lr+1, r);
    return node;
    }

  • atul

    given pre order of BST , it can be formed simply by inserting elements into binary search tree.
    correct me if i am wrong.

     
    node *bst(node *root,int val)
      if(root==NULL)
      {
        temp=createNode(val);
        return temp;
      }
      else if(root->data > val)
      {
         root->left=bst(root->left,val);
      }
      else if(root->data < val)
      {
         root->right=bst(root->right,val);
      }
    
    return root;
    }
     
    • Kartik

      @atul: Thanks for sharing your thoughts. Looks like another working solution. The time complexity of this solution will be O(n^2) for a skewed tree.

      • damon

        @Kartic: i think atual code’s complexity will be atmost nlogn

    • Daniel

      That’s not exactly correct.
      Your solution just creates a BST from any list of elements, but it does not recreate the original BST (like method 1).
      Compare your solution and method 1 on:
      {5, 30, 15, 10, 50, 40, 20}

      • Aaman

        I think Daniel is right this method looks wrong to me ,there is no way to check if a bigger value is suppose to come to current node right or the parent nodes right.

         
        /* Paste your code here (You may delete these lines if not writing code) */