Tree from Postorder and Inorder

Problem Statement - link #

Given inorder and postorder traversals of a Binary Tree in the arrays in[] and post[] respectively. The task is to construct the binary tree from these traversals.

Your Task: You do not need to read input or print anything. Complete the function buildTree() which takes the inorder, postorder traversals and the number of nodes in the tree as input parameters and returns the root node of the newly constructed Binary Tree. The generated output contains the preorder traversal of the constructed tree.

Expected Time Complexity: O(N*N)
Expected Auxiliary Space: O(N)

Examples #

Example 1:

Input:
N = 5
in[] = 9 5 2 3 4
post[] = 5 9 3 4 2
Output: 2 9 5 4 3
Explanation:  
the  resultant binary tree will be
           2
        /     \
       9       4
        \     /
         5   3   

Example 2:

Input:
N = 8
in[] = 4 8 2 5 1 6 3 7
post[] =8 4 5 2 6 7 3 1
Output: 1 2 4 8 5 3 6 7
Explanation: For the given postorder and
inorder traversal of tree the  resultant
binary tree will be
           1
       /      \
     2         3
   /  \      /  \
  4    5    6    7
   \
     8

Constraints #

• 1 <= Number of nodes <= 10^3
• 1 <= Number of elements in inorder and postorder arrays <= 10^3

Solutions #

int posIndex;
Node* cbt(int in[], int post[], int is, int ie){
    if(is>ie) return NULL;
    Node* root = new Node(post[posIndex--]);
    int i = is;
    while(i<=ie && in[i]!=root->data) i++;
    root->right = cbt(in,post,i+1,ie);
    root->left = cbt(in,post,is,i-1);
    return root;
}
//Function to return a tree created from postorder and inoreder traversals.
Node *buildTree(int in[], int post[], int n) {
    // Your code here
    posIndex = n-1;
    return cbt(in,post,0,n-1);
}