对由值为 1 的节点组成的二叉树中的级别进行计数，这些节点组合在一起
原文:https://www . geesforgeks . org/count-levels-in-a-a-binary-tree-由节点组成-value-1-group-together/
给定一个仅由 0 s 和 1 s 组成的二叉树，任务是打印二叉树中的级别计数，其中所有 1 s 连续放置在一个组中。
示例:
输入: 0
/ \
1 0
/ \ / \
1 0 1 0
输出: 2
说明:在 1 级和 2 级中，值为 1 的所有节点都是连续放置的。
输入: 0
/ \
1 0
/ \ \
1 1 0
/ \ \ / \
1 1 1 0 0
输出: 4
说明:在所有级别中，值为 1 的节点是连续放置的。
方法:按照以下步骤解决问题:
- 使用队列执行级顺序遍历。
- 遍历二叉树的每一级，考虑以下三个变量:
- 第一次出现值为 1 的节点后，标志 1: 设置为 1 。
- 第一次出现值为 0 的节点后设置为 1 ，第二次出现值为 1 的节点后设置为。
- 标志 2: 在第一次出现值为 1 的节点后设置，此时标志 0 和标志 1 都设置为 1 。
- 遍历每个级别后，检查标志 1 是否设置为 1，标志 2 是否为 0。如果发现是真的，将该级别包括在计数中。
- 最后，打印获得的计数。
下面是上述方法的实现:
C++
// C++ Program to implement
// the above approach
#include <bits/stdc++.h>
using namespace std;
// A Binary Tree Node
struct node {
// Left Child
struct node* left;
int data;
// Right Child
struct node* right;
};
// Function to perform level order traversal
// count levels with all 1s grouped together
int countLevels(node* root)
{
if (root == NULL)
return 0;
int Count = 0;
// Create an empty queue for
// level order traversal
queue<node*> q;
// Stores front element of the queue
node* curr;
// Enqueue root and NULL node
q.push(root);
// Stores Nodes of
// Current Level
while (!q.empty()) {
int n = q.size();
int flag0 = 0, flag1 = 0, flag2 = 0;
while (n--) {
// Stores first node of
// the current level
curr = q.front();
q.pop();
if (curr) {
// If left child exists
if (curr->left)
// Push into the Queue
q.push(curr->left);
// If right child exists
if (curr->right)
// Push into the Queue
q.push(curr->right);
if (curr->data == 1) {
// If current node is the first
// node with value 1
if (!flag1)
flag1 = 1;
// If current node has value 1
// after occurrence of nodes
// with value 0 following a
// sequence of nodes with value 1
if (flag1 && flag0)
flag2 = 1;
}
// If current node is the first node
// with value 0 after a sequence
// of nodes with value 1
if (curr->data == 0 && flag1)
flag0 = 1;
}
}
if (flag1 && !flag2)
Count++;
}
return Count;
}
// Function to create a Tree Node
node* newNode(int data)
{
node* temp = new node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Driver Code
int main()
{
node* root = newNode(0);
root->left = newNode(0);
root->right = newNode(1);
root->left->left = newNode(0);
root->left->right = newNode(1);
root->right->left = newNode(1);
root->right->right = newNode(0);
cout << countLevels(root);
return 0;
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java Program to implement
// the above approach
import java.util.*;
class GFG
{
// A Binary Tree Node
static class node
{
// Left Child
node left;
int data;
// Right Child
node right;
};
// Function to perform level order traversal
// count levels with all 1s grouped together
static int countLevels(node root)
{
if (root == null)
return 0;
int Count = 0;
// Create an empty queue for
// level order traversal
Queue<node> q = new LinkedList<>();
// Stores front element of the queue
node curr;
// Enqueue root and null node
q.add(root);
// Stores Nodes of
// Current Level
while (!q.isEmpty())
{
int n = q.size();
int flag0 = 0, flag1 = 0, flag2 = 0;
while (n-- >0)
{
// Stores first node of
// the current level
curr = q.peek();
q.remove();
if (curr != null)
{
// If left child exists
if (curr.left != null)
// Push into the Queue
q.add(curr.left);
// If right child exists
if (curr.right != null)
// Push into the Queue
q.add(curr.right);
if (curr.data == 1)
{
// If current node is the first
// node with value 1
if (flag1 == 0)
flag1 = 1;
// If current node has value 1
// after occurrence of nodes
// with value 0 following a
// sequence of nodes with value 1
if (flag1 > 0 && flag0 > 0)
flag2 = 1;
}
// If current node is the first node
// with value 0 after a sequence
// of nodes with value 1
if (curr.data == 0 && flag1 > 0)
flag0 = 1;
}
}
if (flag1 > 0 && flag2 == 0)
Count++;
}
return Count;
}
// Function to create a Tree Node
static node newNode(int data)
{
node temp = new node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
// Driver Code
public static void main(String[] args)
{
node root = newNode(0);
root.left = newNode(0);
root.right = newNode(1);
root.left.left = newNode(0);
root.left.right = newNode(1);
root.right.left = newNode(1);
root.right.right = newNode(0);
System.out.print(countLevels(root));
}
}
// This code is contributed by 29AjayKumar
Python 3
# Python3 program to implement
# the above approach
from collections import deque
# A Binary Tree Node
class node:
def __init__(self):
# Left Child
self.left = None
self.data = 0
# Right Child
self.right = None
# Function to perform level order traversal
# count levels with all 1s grouped together
def countLevels(root):
if (root == None):
return 0
Count = 0
# Create an empty queue for
# level order traversal
q = deque()
# Stores front element of the queue
curr = node()
# Enqueue root and None node
q.append(root)
# Stores Nodes of
# Current Level
while q:
n = len(q)
flag0 = 0
flag1 = 0
flag2 = 0
while (n):
# Stores first node of
# the current level
curr = q[0]
q.popleft()
if (curr):
# If left child exists
if (curr.left):
# Push into the Queue
q.append(curr.left)
# If right child exists
if (curr.right):
# Push into the Queue
q.append(curr.right)
if (curr.data == 1):
# If current node is the first
# node with value 1
if (not flag1):
flag1 = 1
# If current node has value 1
# after occurrence of nodes
# with value 0 following a
# sequence of nodes with value 1
if (flag1 and flag0):
flag2 = 1
# If current node is the first node
# with value 0 after a sequence
# of nodes with value 1
if (curr.data == 0 and flag1):
flag0 = 1
n -= 1
if (flag1 and not flag2):
Count += 1
return Count
# Function to create a Tree Node
def newNode(data):
temp = node()
temp.data = data
temp.left = None
temp.right = None
return temp
# Driver Code
if __name__ == "__main__":
root = newNode(0)
root.left = newNode(0)
root.right = newNode(1)
root.left.left = newNode(0)
root.left.right = newNode(1)
root.right.left = newNode(1)
root.right.right = newNode(0)
print(countLevels(root))
# This code is contributed by sanjeev2552
C
// C# Program to implement
// the above approach
using System;
using System.Collections.Generic;
class GFG
{
// A Binary Tree Node
class node
{
// Left Child
public node left;
public int data;
// Right Child
public node right;
};
// Function to perform level order traversal
// count levels with all 1s grouped together
static int countLevels(node root)
{
if (root == null)
return 0;
int Count = 0;
// Create an empty queue for
// level order traversal
Queue<node> q = new Queue<node>();
// Stores front element of the queue
node curr;
// Enqueue root and null node
q.Enqueue(root);
// Stores Nodes of
// Current Level
while (q.Count != 0)
{
int n = q.Count;
int flag0 = 0, flag1 = 0, flag2 = 0;
while (n-- >0)
{
// Stores first node of
// the current level
curr = q.Peek();
q.Dequeue();
if (curr != null)
{
// If left child exists
if (curr.left != null)
// Push into the Queue
q.Enqueue(curr.left);
// If right child exists
if (curr.right != null)
// Push into the Queue
q.Enqueue(curr.right);
if (curr.data == 1)
{
// If current node is the first
// node with value 1
if (flag1 == 0)
flag1 = 1;
// If current node has value 1
// after occurrence of nodes
// with value 0 following a
// sequence of nodes with value 1
if (flag1 > 0 && flag0 > 0)
flag2 = 1;
}
// If current node is the first node
// with value 0 after a sequence
// of nodes with value 1
if (curr.data == 0 && flag1 > 0)
flag0 = 1;
}
}
if (flag1 > 0 && flag2 == 0)
Count++;
}
return Count;
}
// Function to create a Tree Node
static node newNode(int data)
{
node temp = new node();
temp.data = data;
temp.left = null;
temp.right = null;
return temp;
}
// Driver Code
public static void Main(String[] args)
{
node root = newNode(0);
root.left = newNode(0);
root.right = newNode(1);
root.left.left = newNode(0);
root.left.right = newNode(1);
root.right.left = newNode(1);
root.right.right = newNode(0);
Console.Write(countLevels(root));
}
}
// This code is contributed by 29AjayKumar
java 描述语言
<script>
// JavaScript program for above approach
// A Binary Tree Node
class node
{
constructor(data) {
this.left = null;
this.right = null;
this.data = data;
}
}
// Function to create a Tree Node
function newNode(data)
{
let temp = new node(data);
return temp;
}
// Function to perform level order traversal
// count levels with all 1s grouped together
function countLevels(root)
{
if (root == null)
return 0;
let Count = 0;
// Create an empty queue for
// level order traversal
let q = [];
// Stores front element of the queue
let curr;
// Enqueue root and null node
q.push(root);
// Stores Nodes of
// Current Level
while (q.length > 0)
{
let n = q.length;
let flag0 = 0, flag1 = 0, flag2 = 0;
while (n-- >0)
{
// Stores first node of
// the current level
curr = q[0];
q.shift();
if (curr != null)
{
// If left child exists
if (curr.left != null)
// Push into the Queue
q.push(curr.left);
// If right child exists
if (curr.right != null)
// Push into the Queue
q.push(curr.right);
if (curr.data == 1)
{
// If current node is the first
// node with value 1
if (flag1 == 0)
flag1 = 1;
// If current node has value 1
// after occurrence of nodes
// with value 0 following a
// sequence of nodes with value 1
if (flag1 > 0 && flag0 > 0)
flag2 = 1;
}
// If current node is the first node
// with value 0 after a sequence
// of nodes with value 1
if (curr.data == 0 && flag1 > 0)
flag0 = 1;
}
}
if (flag1 > 0 && flag2 == 0)
Count++;
}
return Count;
}
let root = newNode(0);
root.left = newNode(0);
root.right = newNode(1);
root.left.left = newNode(0);
root.left.right = newNode(1);
root.right.left = newNode(1);
root.right.right = newNode(0);
document.write(countLevels(root));
</script>
Output:
2
时间复杂度:O(N) T5辅助空间:** O(N)
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