buildBinaryTreeFromInorderAndPostOrder

/************************************************************
  
   Following is the TreeNode class structure
   
   class TreeNode<T> { 
        T data; 
        TreeNode<T> left;
        TreeNode<T> right;
  
        TreeNode(T data) { 
            this.data = data; 
            left = null; 
            right = null; 
        }
   }
    
 ************************************************************/

public class Solution {
	public static TreeNode<Integer> getTreeFromPostorderAndInorder(int[] postOrder, int[] inOrder) {
		return buildTree(postOrder, inOrder, 0, inOrder.length - 1);
	}
    
    public static TreeNode<Integer> buildTree(int[] post, int[] in, int s, int e){ // s == > start , e ==> end
        
        if(s > e){
            return null;
        }
        TreeNode<Integer> root = new TreeNode(post[post.length - 1]);
        if(s == e){
            return root;
        }
        int k = 0;
        int i = s;
        // find number of value in left and right subtree
        
        while(in[i] != post[post.length - 1]){
            k++;
            i++;
        }
        
        int leftSubTree[] = new int[k];
        int rightSubTree[] = new int[e - s - k];
        int  l = 0, r = 0;
         for(int j = 0; j < e - s; j++){
             if(k-- > 0)
                leftSubTree[l++] = post[j];
             else
                rightSubTree[r++] = post[j];
        }
     
        
        root.left = buildTree(leftSubTree, in, s, i - 1);
        root.right = buildTree(rightSubTree, in, i + 1, e);
        return root;
        
    }
    
    
}