Construct a Binary Search Tree from given postorder

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

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

     10
   /   \
  5     40
 /  \      \
1    7      50

Method 1 ( O(n^2) time complexity )
The last element of postorder traversal is always root. We first construct the root. Then we find the index of last element which is smaller than root. Let the index be ‘i’. The values between 0 and ‘i’ are part of left subtree, and the values between ‘i+1’ and ‘n-2’ are part of right subtree. Divide given post[] at index “i” and recur for left and right sub-trees.
For example in {1, 7, 5, 40, 50, 10}, 10 is the last element, so we make it root. Now we look for the last element smaller than 10, we find 5. So we know the structure of BST is as following.

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

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

Method 2 ( O(n) time complexity )
The trick is to set a range {min .. max} for every node. Initialize the range as {INT_MIN .. INT_MAX}. The last 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 .. INT_MAX}.

Following code is used to generate the exact Binary Search Tree of a given post order traversal.

• C

/* A O(n) program for construction of BST from    postorder 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, *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 post[]. // postIndex is used to keep track of index in post[]. struct node* constructTreeUtil(int post[], int* postIndex,                          int key, int min, int max, int size) {     // Base case     if (*postIndex < 0)         return NULL;     struct node* root = NULL;     // If current element of post[] is in range, then     // only it is part of current subtree     if (key > min && key < max)     {         // Allocate memory for root of this subtree and decrement         // *postIndex         root = newNode(key);         *postIndex = *postIndex - 1;         if (*postIndex >= 0)         {           // 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(post, postIndex, post[*postIndex],                                           key, max, 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(post, postIndex, post[*postIndex],                                          min, key, size );         }     }     return root; } // The main function to construct BST from given postorder // traversal. This function mainly uses constructTreeUtil() struct node *constructTree (int post[], int size) {     int postIndex = size-1;     return constructTreeUtil(post, &postIndex, post[postIndex],                              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 post[] = {1, 7, 5, 50, 40, 10};     int size = sizeof(post) / sizeof(post[0]);     struct node *root = constructTree(post, size);     printf("Inorder traversal of the constructed tree: \n");     printInorder(root);     return 0; }

Output:

Inorder traversal of the constructed tree: 
1 5 7 10 40 50 

Note that the output to the program will always be a sorted sequence as we are printing the inorder traversal of a Binary Search Tree.