Root to leaf path sum equal to a given number | A binary tree node has data , pointer to left child and a pointer to right child ; Given a tree and a sum , return true if there is a path from the root down to a leaf , such that adding up all the values along the path equals the given sum . Strategy : subtract the node value from the sum when recurring down , and check to see if the sum is 0 when you run out of tree . ; Driver Code ; Constructed binary tree is 10 / \ 8 2 / \ / 3 5 2
class Node { int data ; Node left , right ; Node ( int item ) { data = item ; left = right = null ; } } class BinaryTree { Node root ; boolean hasPathSum ( Node node , int sum ) { if ( node == null ) return sum == 0 ; return hasPathSum ( node . left , sum - node . data ) || hasPathSum ( node . right , sum - node . data ) ; } public static void main ( String args [ ] ) { int sum = 21 ; BinaryTree tree = new BinaryTree ( ) ; tree . root = new Node ( 10 ) ; tree . root . left = new Node ( 8 ) ; tree . root . right = new Node ( 2 ) ; tree . root . left . left = new Node ( 3 ) ; tree . root . left . right = new Node ( 5 ) ; tree . root . right . left = new Node ( 2 ) ; if ( tree . haspathSum ( tree . root , sum ) ) System . out . println ( " There ▁ is ▁ a ▁ root ▁ to ▁ leaf ▁ path ▁ with ▁ sum ▁ " + sum ) ; else System . out . println ( " There ▁ is ▁ no ▁ root ▁ to ▁ leaf ▁ path ▁ with ▁ sum ▁ " + sum ) ; } }