text: Constructed binary tree is 10 / \ 8 2 / \ / 3 5 2
code: struct node * root = newnode ( 10 ) ; root -> left = newnode ( 8 ) ; root -> right = newnode ( 2 ) ; root -> left -> left = newnode ( 3 ) ; root -> left -> right = newnode ( 5 ) ; root -> right -> left = newnode ( 2 ) ; if ( hasPathSum ( root , sum ) ) printf ( " There ▁ is ▁ a ▁ root - to - leaf ▁ path ▁ with ▁ sum ▁ % d " , sum ) ; else printf ( " There ▁ is ▁ no ▁ root - to - leaf ▁ path ▁ with ▁ sum ▁ % d " , sum ) ; getchar ( ) ; return 0 ; }