📜  检查给定树的左视图是否已排序

📅  最后修改于: 2021-09-02 06:32:04             🧑  作者: Mango

给定一棵树,我们的任务是检查它的左视图是否已排序。如果是,则返回true否则返回false。

例子:

方法:

为了解决上面提到的问题,我们必须在树上执行层序遍历并寻找每一层的第一个节点。然后初始化一个变量以检查其左视图是否已排序。如果它没有排序,那么我们可以中断循环并打印 false else 循环继续,最后打印 true。

下面是上述方法的实现:

C++
// C++ implementation to Check if the Left
// View of the given tree is Sorted or not
  
#include 
using namespace std;
  
// Binary Tree Node
struct node {
    int val;
    struct node *right, *left;
};
  
// Utility function to create a new node
struct node* newnode(int key)
{
    struct node* temp = new node;
    temp->val = key;
    temp->right = NULL;
    temp->left = NULL;
  
    return temp;
}
  
// Fucntion to find left view
// and check if it is sorted
void func(node* root)
{
    // queue to hold values
    queue q;
  
    // variable to check whether
    // level order is sorted or not
    bool t = true;
    q.push(root);
  
    int i = -1, j = -1, k = -1;
  
    // Iterate until the queue is empty
    while (!q.empty()) {
        int h = q.size();
  
        // Traverse every level in tree
        while (h > 0) {
            root = q.front();
  
            // variable for initial level
            if (i == -1) {
                j = root->val;
            }
            // checking values are sorted or not
            if (i == -2) {
                if (j <= root->val) {
                    j = root->val;
                    i = -3;
                }
                else {
                    t = false;
                    break;
                }
            }
            // Push left value if it is not null
            if (root->left != NULL) {
                q.push(root->left);
            }
            // Push right value if it is not null
            if (root->right != NULL) {
                q.push(root->right);
            }
            h = h - 1;
  
            // Pop out the values from queue
            q.pop();
        }
        i = -2;
  
        // Check if the value are not
        // sorted then break the loop
        if (t == false) {
            break;
        }
    }
    if (t)
        cout << "true" << endl;
  
    else
        cout << "false" << endl;
}
  
// Driver code
int main()
{
    struct node* root = newnode(10);
    root->right = newnode(50);
    root->right->right = newnode(15);
    root->left = newnode(20);
    root->left->left = newnode(50);
    root->left->right = newnode(23);
    root->right->left = newnode(10);
  
    func(root);
  
    return 0;
}


Java
// Java implementation to check if the left
// view of the given tree is sorted or not
import java.util.*;
  
class GFG{
  
// Binary Tree Node
static class node 
{
    int val;
    node right, left;
};
  
// Utility function to create a new node
static node newnode(int key)
{
    node temp = new node();
    temp.val = key;
    temp.right = null;
    temp.left = null;
  
    return temp;
}
  
// Function to find left view
// and check if it is sorted
static void func(node root)
{
      
    // Queue to hold values
    Queue q = new LinkedList<>();
  
    // Variable to check whether
    // level order is sorted or not
    boolean t = true;
    q.add(root);
  
    int i = -1, j = -1;
  
    // Iterate until the queue is empty
    while (!q.isEmpty())
    {
        int h = q.size();
  
        // Traverse every level in tree
        while (h > 0)
        {
            root = q.peek();
  
            // Variable for initial level
            if (i == -1)
            {
                j = root.val;
            }
              
            // Checking values are sorted or not
            if (i == -2)
            {
                if (j <= root.val)
                {
                    j = root.val;
                    i = -3;
                }
                else 
                {
                    t = false;
                    break;
                }
            }
              
            // Push left value if it is not null
            if (root.left != null) 
            {
                q.add(root.left);
            }
              
            // Push right value if it is not null
            if (root.right != null) 
            {
                q.add(root.right);
            }
            h = h - 1;
  
            // Pop out the values from queue
            q.remove();
        }
        i = -2;
  
        // Check if the value are not
        // sorted then break the loop
        if (t == false) 
        {
            break;
        }
    }
    if (t)
        System.out.print("true" + "\n");
  
    else
        System.out.print("false" + "\n");
}
  
// Driver code
public static void main(String[] args)
{
    node root = newnode(10);
    root.right = newnode(50);
    root.right.right = newnode(15);
    root.left = newnode(20);
    root.left.left = newnode(50);
    root.left.right = newnode(23);
    root.right.left = newnode(10);
  
    func(root);
}
}
  
// This code is contributed by gauravrajput1


C#
// C# implementation to check if the left
// view of the given tree is sorted or not
using System;
using System.Collections.Generic;
  
class GFG{
  
// Binary Tree Node
class node 
{
    public int val;
    public node right, left;
};
  
// Utility function to create a new node
static node newnode(int key)
{
    node temp = new node();
    temp.val = key;
    temp.right = null;
    temp.left = null;
  
    return temp;
}
  
// Function to find left view
// and check if it is sorted
static void func(node root)
{
      
    // Queue to hold values
    Queue q = new Queue();
  
    // Variable to check whether
    // level order is sorted or not
    bool t = true;
    q.Enqueue(root);
  
    int i = -1, j = -1;
  
    // Iterate until the queue is empty
    while (q.Count != 0)
    {
        int h = q.Count;
  
        // Traverse every level in tree
        while (h > 0)
        {
            root = q.Peek();
  
            // Variable for initial level
            if (i == -1)
            {
                j = root.val;
            }
              
            // Checking values are sorted or not
            if (i == -2)
            {
                if (j <= root.val)
                {
                    j = root.val;
                    i = -3;
                }
                else
                {
                    t = false;
                    break;
                }
            }
              
            // Push left value if it is not null
            if (root.left != null) 
            {
                q.Enqueue(root.left);
            }
              
            // Push right value if it is not null
            if (root.right != null) 
            {
                q.Enqueue(root.right);
            }
            h = h - 1;
  
            // Pop out the values from queue
            q.Dequeue();
        }
        i = -2;
  
        // Check if the value are not
        // sorted then break the loop
        if (t == false) 
        {
            break;
        }
    }
    if (t)
        Console.Write("true" + "\n");
  
    else
        Console.Write("false" + "\n");
}
  
// Driver code
public static void Main(String[] args)
{
    node root = newnode(10);
    root.right = newnode(50);
    root.right.right = newnode(15);
    root.left = newnode(20);
    root.left.left = newnode(50);
    root.left.right = newnode(23);
    root.right.left = newnode(10);
  
    func(root);
}
}
  
// This code is contributed by gauravrajput1


输出:
true

时间复杂度: O(N)

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