计算二叉树奇数层节点和偶数层节点之和之差的 Java 程序

原文:https://www . geesforgeks . org/Java-计算二叉树奇数层和偶数层节点之和的程序差/

使用 DFS 的图遍历是一种通过递归遍历树的显而易见的方法。下面是一个使用 DFS 遍历二叉树的算法。

先决条件

算法

  1. 将当前节点初始化为根节点，将父节点初始化为-1。
  2. 按照一般的 DFS 方法遍历二叉树，随着我们遍历距离根节点越远，节点的级别越高。
  3. 遍历时，我们检查二叉树当前节点的级别是否为偶数，然后添加偶数和，否则添加奇数和。
  4. 最后打印偶数和奇数和的绝对差。

示例

Java

import java.util.*;
public class GFG {

    // global variable declaration
    static ArrayList<ArrayList<Integer> > 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 recurssion
        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));
    }
}

输出

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

时间复杂度:O(V+E)V为顶点 E 为边。

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