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# Java Program to Calculate the Difference Between the Sum of the Odd Level and the Even Level Nodes of a Binary Tree

Graph Traversal using DFS is an obvious way to traverse a tree with recursion. Below is an algorithm for traversing binary tree using DFS.

Prerequisites

Algorithm

1. Initialize the current node as root node and the parent as -1.
2. Traverse the Binary Tree as the in the general DFS fashion and keep of increasing the level of the node as we traverse farther from the root node.
3. While traversing we check if the level of the current node of the binary tree is even then add in even sum else add in odd sum.
4. Finally, print the Absolute difference of the of even sum and the odd sum.

Example

## Java

 `import` `java.util.*;` `public` `class` `GFG {` `  `  `    ``// global variable declaration` `    ``static` `ArrayList > arr;` `    ``static` `int` `val[];` `    ``static` `int` `sum_odd = ``0``, sum_even = ``0``;`   `    ``// traverses the binary-tree/tree having parameters u,` `    ``// par, level which denotes current node, current's` `    ``// parent node, current level of the tree.` `    ``static` `void` `dfs(``int` `u, ``int` `par, ``int` `level)` `    ``{` `        ``// according to level adding the node` `        ``if` `(level % ``2` `== ``0``)` `            ``sum_even += val[u];` `        ``else` `            ``sum_odd += val[u];`   `        ``// exploring the child of the particular node u (2` `        ``// in case of binary tree).` `        ``for` `(``int` `v : arr.get(u)) {` `            ``if` `(v != par) {` `              `  `                ``// recursively calling the current child` `                ``// node to become parent of the next dfs` `                ``// call.` `                ``dfs(v, u, level + ``1``);` `            ``}` `        ``}` `    ``}`   `    ``public` `static` `void` `main(String args[])` `    ``{` `        ``Scanner in = ``new` `Scanner(System.in);` `        ``int` `n = ``5``;` `        ``val = ``new` `int``[] { ``0``, ``2``, ``10``, ``5``, ``3``, ``2` `};` `      `  `        ``// declaration of the ArrayList size` `        ``arr = ``new` `ArrayList<>();` `      `  `        ``// initialization of each array element as ArrayList` `        ``// class` `        ``for` `(``int` `i = ``0``; i <= n; i++)` `            ``arr.add(``new` `ArrayList<>());`   `        ``arr.get(``1``).add(``2``);` `        ``arr.get(``2``).add(``1``);`   `        ``arr.get(``1``).add(``4``);` `        ``arr.get(``4``).add(``1``);`   `        ``arr.get(``2``).add(``5``);` `        ``arr.get(``5``).add(``2``);`   `        ``arr.get(``3``).add(``4``);` `        ``arr.get(``4``).add(``3``);`   `        ``//         1(2)` `        ``//    /     \` `        ``//   2(10)     4(3)` `        ``//  /         /` `        ``// 5(2)   3(5)`   `        ``// initial call of recursion` `        ``dfs(``1``, -``1``, ``0``);`   `        ``System.out.println(` `            ``"Absolute difference of sum of odd and even nodes of a binary tree "` `            ``+ Math.abs(sum_odd - sum_even));` `    ``}` `}`

Output

`Absolute difference of sum of odd and even nodes of a binary tree 4`

Time Complexity: O(V + E) where V is the vertices and E is the edges.

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