完全二叉树镜像中的边数
给定深度为 H 的完整二叉树。如果从这棵树的左右两边拍摄镜像，那么:
右镜像:每一级的最右边节点连接到镜像对应节点。 左侧镜像图像:每一级的最左侧节点连接到镜像对应的节点。
任务是在最终的树中获取两个镜像后找到边的数量。
例:
输入: H = 1 输出: 10 原始树中的 2 条边将镜像到镜像中(左右)，即总共 6 条边。 以及连接镜像和原始树的边，如上图所示。 输入: H = 2 输出: 24 (6 * 3) + 3 + 3 = 24
方法:保持每个镜像后最左边、最右边的节点。镜像的每次操作后，边数都会改变。最初，
右镜像后:
左镜像后:
在完全修改的树中:
以下是上述方法的实现:
C++
// C++ implementation of the approach
#include <bits/stdc++.h>
using namespace std;
// Function to return the total number
// of edges in the modified tree
int countEdges(int H)
{
int edges, right, left;
edges = 2 * (pow(2, H) - 1);
left = right = H + 1;
// Total edges in the modified tree
int cnt = (edges * 3) + left + right;
return cnt;
}
// Driver code
int main()
{
int H = 1;
cout << countEdges(H);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java implementation of the approach
import java.io.*;
class GFG {
// Function to return the total number
// of edges in the modified tree
static int countEdges(int H)
{
int edges, right, left;
edges = 2 * (int)(Math.pow(2, H) - 1);
left = right = H + 1;
// Total edges in the modified tree
int cnt = (edges * 3) + left + right;
return cnt;
}
// Driver code
public static void main(String[] args)
{
int H = 1;
System.out.println(countEdges(H));
}
}
// This code has been contributed by anuj_67..
Python 3
# Python 3 implementation of the approach
# Function to return the total number
# of edges in the modified tree
def countEdges( H):
edges = 2 * (pow(2, H) - 1)
left = right = H + 1
# Total edges in the modified tree
cnt = (edges * 3) + left + right
return cnt
# Driver code
if __name__ == "__main__":
H = 1;
print(countEdges(H))
# This code is contributed by ChitraNayal
C
// C# implementation of the approach
using System;
class GFG
{
// Function to return the total number
// of edges in the modified tree
static int countEdges(int H)
{
int edges, right, left;
edges = 2 * (int)(Math.Pow(2, H) - 1);
left = right = H + 1;
// Total edges in the modified tree
int cnt = (edges * 3) + left + right;
return cnt;
}
// Driver code
public static void Main()
{
int H = 1;
Console.WriteLine(countEdges(H));
}
}
// This code is contributed by AnkitRai01
java 描述语言
<script>
// Javascript implementation of the approach
// Function to return the total number
// of edges in the modified tree
function countEdges(H)
{
let edges, right, left;
edges = 2 * (Math.pow(2, H) - 1);
left = right = H + 1;
// Total edges in the modified tree
let cnt = (edges * 3) + left + right;
return cnt;
}
let H = 1;
document.write(countEdges(H));
</script>
Output:
10
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