Tree from Postorder and Inorder

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.

 

Example 1:

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

 

Example 2:

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

 

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(N2)
Expected Auxiliary Space: O(N)

 

Constraints:
1 <= N <= 103
1 <= in[i], post[i] <= 10^3

Solution :

/* Tree node structure

struct Node
{
    int data;
    struct Node* left;
    struct Node* right;

    Node(int x){
        data = x;
        left = right = NULL;
    }
};*/

//Function to return a tree created from postorder and inoreder traversals.

Node* newNode(int data);

Node* buildUtil(int in[], int post[], int inStrt,
    int inEnd, int* pIndex, unordered_map<int, int>& mp)
{
    // Base case
    if (inStrt > inEnd)
        return NULL;
 
    /* Pick current node from Postorder traversal 
    using postIndex and decrement postIndex */
    int curr = post[*pIndex];
    Node* node = newNode(curr);
    (*pIndex)--;
 
    /* If this node has no children then return */
    if (inStrt == inEnd)
        return node;
 
    /* Else find the index of this node in Inorder
    traversal */
    int iIndex = mp[curr];
 
    /* Using index in Inorder traversal, construct
    left and right subtress */
    node->right = buildUtil(in, post, iIndex + 1,
                            inEnd, pIndex, mp);
    node->left = buildUtil(in, post, inStrt,
                           iIndex - 1, pIndex, mp);
 
    return node;
}

Node *buildTree(int in[], int post[], int len) {
    unordered_map<int, int> mp;
    for (int i = 0; i < len; i++)
        mp[in[i]] = i;
 
    int index = len - 1; // Index in postorder
    return buildUtil(in, post, 0, len - 1, &index, mp);
}

Node* newNode(int data)
{
    Node* node = (Node*)malloc(sizeof(Node));
    node->data = data;
    node->left = node->right = NULL;
    return (node);
}