给定一棵仅由值为0和1的节点组成的二叉树，任务是找到通过从左到右连接每个级别的同一级别的节点所形成的二进制数的十进制等值的总和。
例子：
Input: Below is the given Tree:
0
/ \
1 0
/ \ / \
0 1 1 1
Output: 9
Explanation:
Binary number formed at level 1 is “0” and its decimal equivalent is 0.
Binary number formed at level 2 is “10” and its decimal equivalent is 2.
Binary number formed at level 3 is “0111” and its decimal equivalent is 7.
Therefore, total sum = 0 + 2 + 7 = 9.
Input: Below is the given Tree:
0
/
1
/ \
1 0
Output: 3
方法：想法是使用队列执行级别顺序遍历并找到在每个级别形成的数字的总和。请按照以下步骤解决问题：
- 初始化一个队列，比如Q ，来执行层序遍历和一个变量，比如ans ，来存储形成的数字的总和。
- 将根节点推入队列Q。
- 迭代直到队列变空并执行以下步骤：
- 迭代范围[1, Q.size()]并连接该级别的所有节点并将相应节点的子节点推送到队列Q 中。
- 找出所形成的二进制数的十进制等价物，并将这个数加到ans 上。
- 完成上述步骤后，打印ans的值作为结果和。
下面是上述方法的实现：
C++
// CPP program for the above approach
#include
using namespace std;
// Structure of a Tree Node
class TreeNode {
public:
int val;
TreeNode *left, *right;
TreeNode(int key) {
val = key;
left = right = NULL;
}
};
// Function to convert binary number
// to its equivalent decimal value
int convertBinaryToDecimal(vector arr) {
int ans = 0;
for (int i : arr) ans = (ans << 1) | i;
return ans;
}
// Function to calculate sum of
// decimal equivalent of binary numbers
// of node values present at each level
void decimalEquilvalentAtEachLevel(TreeNode *root) {
int ans = 0;
queue que;
// Push root node into queue
que.push(root);
while (true) {
int length = que.size();
if (length == 0) break;
vector eachLvl;
// Connect nodes at the same
// level to form a binary number
while (length > 0) {
TreeNode *temp = que.front();
que.pop();
// Append the value of the
// current node to eachLvl
eachLvl.push_back(temp->val);
// Insert the Left child to
// queue, if its not NULL
if (temp->left != NULL) que.push(temp->left);
// Insert the Right child to
// queue, if its not NULL
if (temp->right != NULL) que.push(temp->right);
// Decrement length by one
length -= 1;
// Stores the front
// element of the queue
}
// Add decimal equivalent of the
// binary number formed on the
// current level to ans
ans += convertBinaryToDecimal(eachLvl);
}
// Finally print ans
cout << ans << endl;
}
// Driver Code
int main()
{
// Given Tree
TreeNode *root = new TreeNode(0);
root->left = new TreeNode(1);
root->right = new TreeNode(0);
root->left->left = new TreeNode(0);
root->left->right = new TreeNode(1);
root->right->left = new TreeNode(1);
root->right->right = new TreeNode(1);
// Function Call
decimalEquilvalentAtEachLevel(root);
return 0;
}
// This code is contributed by sanjeev2552 Java
// Java program for the above approach
import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
class GFG {
// Structure of a Tree Node
static class TreeNode {
int val;
TreeNode left, right;
public TreeNode(int key) {
val = key;
left = right = null;
}
}
// Function to convert binary number
// to its equivalent decimal value
static int convertBinaryToDecimal(ArrayList arr) {
int ans = 0;
for (int i : arr)
ans = (ans << 1) | i;
return ans;
}
// Function to calculate sum of
// decimal equivalent of binary numbers
// of node values present at each level
static void decimalEquilvalentAtEachLevel(TreeNode root) {
int ans = 0;
Queue que = new LinkedList<>();
// Push root node into queue
que.add(root);
while (true) {
int length = que.size();
if (length == 0)
break;
ArrayList eachLvl = new ArrayList<>();
// Connect nodes at the same
// level to form a binary number
while (length > 0) {
TreeNode temp = que.poll();
// Append the value of the
// current node to eachLvl
eachLvl.add(temp.val);
// Insert the Left child to
// queue, if its not NULL
if (temp.left != null)
que.add(temp.left);
// Insert the Right child to
// queue, if its not NULL
if (temp.right != null)
que.add(temp.right);
// Decrement length by one
length -= 1;
// Stores the front
// element of the queue
}
// Add decimal equivalent of the
// binary number formed on the
// current level to ans
ans += convertBinaryToDecimal(eachLvl);
}
// Finally print ans
System.out.println(ans);
}
// Driver Code
public static void main(String[] args) {
// Given Tree
TreeNode root = new TreeNode(0);
root.left = new TreeNode(1);
root.right = new TreeNode(0);
root.left.left = new TreeNode(0);
root.left.right = new TreeNode(1);
root.right.left = new TreeNode(1);
root.right.right = new TreeNode(1);
// Function Call
decimalEquilvalentAtEachLevel(root);
}
// This code is contributed by sanjeev2552
} Python3
# Python3 program for the above approach
# Structure of a Tree Node
class TreeNode:
def __init__(self, val = 0,
left = None, right = None):
self.val = val
self.left = left
self.right = right
# Function to convert binary number
# to its equivalent decimal value
def convertBinaryToDecimal(arr):
ans = 0
for i in arr:
ans = (ans << 1) | i
return ans
# Function to calculate sum of
# decimal equivalent of binary numbers
# of node values present at each level
def decimalEquilvalentAtEachLevel(root):
ans = 0
# Push root node into queue
que = [root]
while True:
length = len(que)
if not length:
break
eachLvl = []
# Connect nodes at the same
# level to form a binary number
while length:
# Stores the front
# element of the queue
temp = que.pop(0)
# Append the value of the
# current node to eachLvl
eachLvl.append(temp.val)
# Insert the Left child to
# queue, if its not NULL
if temp.left:
que.append(temp.left)
# Insert the Right child to
# queue, if its not NULL
if temp.right:
que.append(temp.right)
# Decrement length by one
length -= 1
# Add decimal equivalent of the
# binary number formed on the
# current level to ans
ans += convertBinaryToDecimal(eachLvl)
# Finally print ans
print(ans)
# Driver Code
# Given Tree
root = TreeNode(0)
root.left = TreeNode(1)
root.right = TreeNode(0)
root.left.left = TreeNode(0)
root.left.right = TreeNode(1)
root.right.left = TreeNode(1)
root.right.right = TreeNode(1)
# Function Call
decimalEquilvalentAtEachLevel(root)C#
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG{
// Structure of a Tree Node
class TreeNode{
public int val;
public TreeNode left,right;
};
static TreeNode newNode(int key){
TreeNode temp = new TreeNode();
temp.val = key;
temp.left = temp.right = null;
return temp;
}
// Function to convert binary number
// to its equivalent decimal value
static int convertBinaryToDecimal(List arr)
{
int ans = 0;
foreach(int i in arr)
ans = (ans << 1) | i;
return ans;
}
// Function to calculate sum of
// decimal equivalent of binary numbers
// of node values present at each level
static void decimalEquilvalentAtEachLevel(TreeNode root){
int ans = 0;
Queue que = new Queue();
// Push root node into queue
que.Enqueue(root);
while(true){
int length = que.Count;
if (length == 0)
break;
List eachLvl = new List();
// Connect nodes at the same
// level to form a binary number
while(length > 0){
TreeNode temp = que.Peek();
que.Dequeue();
// Append the value of the
// current node to eachLvl
eachLvl.Add(temp.val);
// Insert the Left child to
// queue, if its not NULL
if (temp.left != null)
que.Enqueue(temp.left);
// Insert the Right child to
// queue, if its not NULL
if (temp.right!=null)
que.Enqueue(temp.right);
// Decrement length by one
length -= 1;
// Stores the front
// element of the queue
}
// Add decimal equivalent of the
// binary number formed on the
// current level to ans
ans += convertBinaryToDecimal(eachLvl);
}
// Finally print ans
Console.WriteLine(ans);
}
// Driver Code
public static void Main()
{
// Given Tree
TreeNode root = newNode(0);
root.left = newNode(1);
root.right = newNode(0);
root.left.left = newNode(0);
root.left.right = newNode(1);
root.right.left = newNode(1);
root.right.right = newNode(1);
// Function Call
decimalEquilvalentAtEachLevel(root);
}
}
// This code is contributed by SURENDRA_GANGWAR. 输出：
9时间复杂度： O(N)
辅助空间： O(1)
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