Second Binary Tree 5 / \ 1 8 \ \ 2 8
Node * root2 = newNode ( 5 ) ; root2 -> left = newNode ( 1 ) ; root2 -> right = newNode ( 8 ) ; root2 -> left -> right = newNode ( 2 ) ; root2 -> right -> right = newNode ( 8 ) ; Node * root3 = MaximumBinaryTree ( root1 , root2 ) ; inorder ( root3 ) ; }