text: 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 . ; Helper function that allocates a new node with the given data and NULL left and right pointers . ; Driver Code ; Constructed binary tree is 10 / \ 8 2 / \ / 3 5 2
code: #include <stdio.h> NEW_LINE #include <stdlib.h> NEW_LINE #define bool  int NEW_LINE struct node { int data ; struct node * left ; struct node * right ; } ; bool hasPathSum ( struct node * node , int sum ) { if ( node == NULL ) { return ( sum == 0 ) ; } else { bool ans = 0 ; int subSum = sum - node -> data ; if ( subSum == 0 && node -> left == NULL && node -> right == NULL ) return 1 ; if ( node -> left ) ans = ans || hasPathSum ( node -> left , subSum ) ; if ( node -> right ) ans = ans || hasPathSum ( node -> right , subSum ) ; return ans ; } } struct node * newnode ( int data ) { struct node * node = ( struct node * ) malloc ( sizeof ( struct node ) ) ; node -> data = data ; node -> left = NULL ; node -> right = NULL ; return ( node ) ; } int main ( ) { int sum = 21 ; 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 ; }