Write a function that returns true if the given Binary Tree is SumTree else false. A SumTree is a Binary Tree where the value of a node is equal to sum of the nodes present in its left subtree and right subtree. An empty tree is SumTree and sum of an empty tree can be considered as 0. A leaf node is also considered as SumTree.
Following is an example of SumTree.
26
/ \
10 3
/ \ \
4 6 3
Method 1 ( Simple )
Get the sum of nodes in left subtree and right subtree. Check if the sum calculated is equal to root’s data. Also, recursively check if the left and right subtrees are SumTrees.
C
#include <stdio.h> #include <stdlib.h> /* A binary tree node has data, left child and right child */struct node { int data; struct node* left; struct node* right; }; /* A utility function to get the sum of values in tree with root as root */int sum(struct node *root) { if(root == NULL) return 0; return sum(root->left) + root->data + sum(root->right); } /* returns 1 if sum property holds for the given node and both of its children */int isSumTree(struct node* node) { int ls, rs; /* If node is NULL or it's a leaf node then return true */ if(node == NULL || (node->left == NULL && node->right == NULL)) return 1; /* Get sum of nodes in left and right subtrees */ ls = sum(node->left); rs = sum(node->right); /* if the node and both of its children satisfy the property return 1 else 0*/ if((node->data == ls + rs)&& isSumTree(node->left) && isSumTree(node->right)) return 1; return 0; } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */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); } /* Driver program to test above function */int main() { struct node *root = newNode(26); root->left = newNode(10); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(6); root->right->right = newNode(3); if(isSumTree(root)) printf("The given tree is a SumTree "); else printf("The given tree is not a SumTree "); getchar(); return 0; } |
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Java
// Java program to check if Binary tree is sum tree or not /* A binary tree node has data, left child and right child */class Node { int data; Node left, right, nextRight; Node(int item) { data = item; left = right = nextRight = null; } } class BinaryTree { Node root; /* A utility function to get the sum of values in tree with root as root */ int sum(Node node) { if (node == null) return 0; return sum(node.left) + node.data + sum(node.right); } /* returns 1 if sum property holds for the given node and both of its children */ int isSumTree(Node node) { int ls, rs; /* If node is NULL or it's a leaf node then return true */ if ((node == null) || (node.left == null && node.right == null)) return 1; /* Get sum of nodes in left and right subtrees */ ls = sum(node.left); rs = sum(node.right); /* if the node and both of its children satisfy the property return 1 else 0*/ if ((node.data == ls + rs) && (isSumTree(node.left) != 0) && (isSumTree(node.right)) != 0) return 1; return 0; } /* Driver program to test above functions */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(26); tree.root.left = new Node(10); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(6); tree.root.right.right = new Node(3); if (tree.isSumTree(tree.root) != 0) System.out.println("The given tree is a sum tree"); else System.out.println("The given tree is not a sum tree"); } } // This code has been contributed by Mayank Jaiswal |
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Output:
The given tree is a SumTree
Time Complexity: O(n^2) in worst case. Worst case occurs for a skewed tree.
Method 2 ( Tricky )
The Method 1 uses sum() to get the sum of nodes in left and right subtrees. The method 2 uses following rules to get the sum directly.
1) If the node is a leaf node then sum of subtree rooted with this node is equal to value of this node.
2) If the node is not a leaf node then sum of subtree rooted with this node is twice the value of this node (Assuming that the tree rooted with this node is SumTree).
C
#include <stdio.h> #include <stdlib.h> /* A binary tree node has data, left child and right child */struct node { int data; struct node* left; struct node* right; }; /* Utillity function to check if the given node is leaf or not */int isLeaf(struct node *node) { if(node == NULL) return 0; if(node->left == NULL && node->right == NULL) return 1; return 0; } /* returns 1 if SumTree property holds for the given tree */int isSumTree(struct node* node) { int ls; // for sum of nodes in left subtree int rs; // for sum of nodes in right subtree /* If node is NULL or it's a leaf node then return true */ if(node == NULL || isLeaf(node)) return 1; if( isSumTree(node->left) && isSumTree(node->right)) { // Get the sum of nodes in left subtree if(node->left == NULL) ls = 0; else if(isLeaf(node->left)) ls = node->left->data; else ls = 2*(node->left->data); // Get the sum of nodes in right subtree if(node->right == NULL) rs = 0; else if(isLeaf(node->right)) rs = node->right->data; else rs = 2*(node->right->data); /* If root's data is equal to sum of nodes in left and right subtrees then return 1 else return 0*/ return(node->data == ls + rs); } return 0; } /* Helper function that allocates a new node with the given data and NULL left and right pointers. */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); } /* Driver program to test above function */int main() { struct node *root = newNode(26); root->left = newNode(10); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(6); root->right->right = newNode(3); if(isSumTree(root)) printf("The given tree is a SumTree "); else printf("The given tree is not a SumTree "); getchar(); return 0; } |
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Java
// Java program to check if Binary tree is sum tree or not /* A binary tree node has data, left child and right child */class Node { int data; Node left, right, nextRight; Node(int item) { data = item; left = right = nextRight = null; } } class BinaryTree { Node root; /* Utility function to check if the given node is leaf or not */ int isLeaf(Node node) { if (node == null) return 0; if (node.left == null && node.right == null) return 1; return 0; } /* returns 1 if SumTree property holds for the given tree */ int isSumTree(Node node) { int ls; // for sum of nodes in left subtree int rs; // for sum of nodes in right subtree /* If node is NULL or it's a leaf node then return true */ if (node == null || isLeaf(node) == 1) return 1; if (isSumTree(node.left) != 0 && isSumTree(node.right) != 0) { // Get the sum of nodes in left subtree if (node.left == null) ls = 0; else if (isLeaf(node.left) != 0) ls = node.left.data; else ls = 2 * (node.left.data); // Get the sum of nodes in right subtree if (node.right == null) rs = 0; else if (isLeaf(node.right) != 0) rs = node.right.data; else rs = 2 * (node.right.data); /* If root's data is equal to sum of nodes in left and right subtrees then return 1 else return 0*/ if ((node.data == rs + ls)) return 1; else return 0; } return 0; } /* Driver program to test above functions */ public static void main(String args[]) { BinaryTree tree = new BinaryTree(); tree.root = new Node(26); tree.root.left = new Node(10); tree.root.right = new Node(3); tree.root.left.left = new Node(4); tree.root.left.right = new Node(6); tree.root.right.right = new Node(3); if (tree.isSumTree(tree.root) != 0) System.out.println("The given tree is a sum tree"); else System.out.println("The given tree is not a sum tree"); } } // This code has been contributed by Mayank Jaiswal |
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Output:
The given tree is a sum tree
Time Complexity: O(n)
Please write comments if you find the above codes/algorithms incorrect, or find other ways to solve the same problem.
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