逐行层序遍历|设置 2（使用两个队列）

给定一个二叉树，逐级打印节点，每个级别在一个新行上。

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Output:
1
2 3
4 5

我们在下面的文章中讨论了一种解决方案。
逐行打印级别顺序遍历|设置 1
在这篇文章中，讨论了使用两个队列的不同方法。我们可以在第一个队列中插入第一级并打印它，在从第一个队列弹出时将其左右节点插入到第二个队列中。现在开始打印第二个队列，在弹出之前将其左右子节点插入第一个队列。继续这个过程，直到两个队列都为空。

C++

// C++ program to do level order traversal line by
// line
#include 
using namespace std;
 
struct Node
{
    int data;
    Node *left, *right;
};
 
// Prints level order traversal line by line
// using two queues.
void levelOrder(Node *root)
{
    queue q1, q2;
 
    if (root == NULL)
        return;
 
    // Pushing first level node into first queue
    q1.push(root);
 
    // Executing loop till both the queues
    // become empty
    while (!q1.empty() || !q2.empty())
    {
        while (!q1.empty())
        {
            // Pushing left child of current node in
            // first queue into second queue
            if (q1.front()->left != NULL)
                q2.push(q1.front()->left);
 
            // pushing right child of current node
            // in first queue into second queue
            if (q1.front()->right != NULL)
                q2.push(q1.front()->right);
 
            cout << q1.front()->data << " ";
            q1.pop();
        }
 
        cout << "\n";
 
        while (!q2.empty())
        {
            // pushing left child of current node
            // in second queue into first queue
            if (q2.front()->left != NULL)
                q1.push(q2.front()->left);
 
            // pushing right child of current
            // node in second queue into first queue
            if (q2.front()->right != NULL)
                q1.push(q2.front()->right);
 
            cout << q2.front()->data << " ";
            q2.pop();
        }
 
        cout << "\n";
    }
}
 
// Utility function to create a new tree node
Node* newNode(int data)
{
    Node *temp = new Node;
    temp->data = data;
    temp->left = NULL;
    temp->right = NULL;
    return temp;
}
 
// Driver program to test above functions
int main()
{
    Node *root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
    root->right->right = newNode(6);
 
    levelOrder(root);
    return 0;
}

Java

//Java program to do level order traversal line by
//line
import java.util.LinkedList;
import java.util.Queue;
 
public class GFG
{
    static class Node
    {
        int data;
        Node left;
        Node right;
 
        Node(int data)
        {
            this.data = data;
            left = null;
            right = null;
        }
    }
 
    // Prints level order traversal line by line
    // using two queues.
    static void levelOrder(Node root)
    {
        Queue q1 = new LinkedList();
        Queue q2 = new LinkedList();
 
        if (root == null)
            return;
 
        // Pushing first level node into first queue
        q1.add(root);
 
        // Executing loop till both the queues
        // become empty
        while (!q1.isEmpty() || !q2.isEmpty())
        {
 
            while (!q1.isEmpty())
            {
 
                // Pushing left child of current node in
                // first queue into second queue
                if (q1.peek().left != null)
                    q2.add(q1.peek().left);
 
                // pushing right child of current node
                // in first queue into second queue
                if (q1.peek().right != null)
                    q2.add(q1.peek().right);
 
                System.out.print(q1.peek().data + " ");
                q1.remove();
            }
            System.out.println();
 
            while (!q2.isEmpty())
            {
 
                // pushing left child of current node
                // in second queue into first queue
                if (q2.peek().left != null)
                    q1.add(q2.peek().left);
 
                // pushing right child of current
                // node in second queue into first queue
                if (q2.peek().right != null)
                    q1.add(q2.peek().right);
 
                System.out.print(q2.peek().data + " ");
                q2.remove();
            }
            System.out.println();
        }
    }
 
    // Driver program to test above functions
    public static void main(String[] args)
    {
 
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.right = new Node(6);
 
        levelOrder(root);
    }
}
// This code is Contributed by Sumit Ghosh

Python

"""
Python program to do level order traversal
line by line using dual queue"""
class GFG:
     
    """Constructor to create a new tree node"""
    def __init__(self,data):
        self.val = data
        self.left = None
        self.right = None
     
    """Prints level order traversal line by
    line using two queues."""
    def levelOrder(self,node):
        q1 = [] # Queue 1
        q2 = [] # Queue 2
        q1.append(node)
         
        """Executing loop till both the
        queues become empty"""
        while(len(q1) > 0 or len(q2) > 0):
         
            """Empty string to concatenate
            the string for q1"""
            concat_str_q1 = ''
            while(len(q1) > 0):
                 
                """Poped node at the first
                pos in queue 1 i.e q1"""
                poped_node = q1.pop(0)
                concat_str_q1 += poped_node.val +' '
                 
                """Pushing left child of current
                node in first queue into second queue"""
                if poped_node.left:
                    q2.append(poped_node.left)
                     
                """Pushing right child of current node
                in first queue into second queue"""
                if poped_node.right:
                    q2.append(poped_node.right)
            print( str(concat_str_q1))
            concat_str_q1 = ''
             
            """Empty string to concatenate the
            string for q1"""
            concat_str_q2 = ''
            while (len(q2) > 0):
             
                """Poped node at the first pos
                in queue 1 i.e q1"""
                poped_node = q2.pop(0)
                concat_str_q2 += poped_node.val + ' '
                 
                """Pushing left child of current node
                in first queue into first queue"""
                if poped_node.left:
                    q1.append(poped_node.left)
                 
                """Pushing right child of current node
                in first queue into first queue"""
                if poped_node.right:
                    q1.append(poped_node.right)
            print(str(concat_str_q2))
            concat_str_q2 = ''
 
""" Driver program to test above functions"""
node = GFG("1")
node.left = GFG("2")
node.right = GFG("3")
node.left.left = GFG("4")
node.left.right = GFG("5")
node.right.right = GFG("6")
node.levelOrder(node)
 
# This code is contributed by Vaibhav Kumar 12

C#

// C# program to do level order
// traversal line by line
using System;
using System.Collections.Generic;
 
class GFG
{
    public class Node
    {
        public int data;
        public Node left;
        public Node right;
 
        public Node(int data)
        {
            this.data = data;
            left = null;
            right = null;
        }
    }
 
    // Prints level order traversal
    // line by line using two queues.
    public static void levelOrder(Node root)
    {
        LinkedList q1 = new LinkedList();
        LinkedList q2 = new LinkedList();
 
        if (root == null)
        {
            return;
        }
 
        // Pushing first level node
        // into first queue
        q1.AddLast(root);
 
        // Executing loop till both the
        // queues become empty
        while (q1.Count > 0 || q2.Count > 0)
        {
 
            while (q1.Count > 0)
            {
 
                // Pushing left child of current node
                // in first queue into second queue
                if (q1.First.Value.left != null)
                {
                    q2.AddLast(q1.First.Value.left);
                }
 
                // pushing right child of current node
                // in first queue into second queue
                if (q1.First.Value.right != null)
                {
                    q2.AddLast(q1.First.Value.right);
                }
 
                Console.Write(q1.First.Value.data + " ");
                q1.RemoveFirst();
            }
            Console.WriteLine();
 
            while (q2.Count > 0)
            {
 
                // pushing left child of current node
                // in second queue into first queue
                if (q2.First.Value.left != null)
                {
                    q1.AddLast(q2.First.Value.left);
                }
 
                // pushing right child of current
                // node in second queue into first queue
                if (q2.First.Value.right != null)
                {
                    q1.AddLast(q2.First.Value.right);
                }
 
                Console.Write(q2.First.Value.data + " ");
                q2.RemoveFirst();
            }
            Console.WriteLine();
        }
    }
 
    // Driver Code
    public static void Main(string[] args)
    {
        Node root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
        root.right.right = new Node(6);
 
        levelOrder(root);
    }
}
 
// This code is contributed by Shrikant13

Javascript

  
// Javascript program to do level order
// traversal line by line
class Node
{
    constructor(data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
}
 
// Prints level order traversal
// line by line using two queues.
function levelOrder(root)
{
    var q1 = [];
    var q2 = [];
    if (root == null)
    {
        return;
    }
     
    // Pushing first level node
    // into first queue
    q1.push(root);
     
    // Executing loop till both the
    // queues become empty
    while (q1.length > 0 || q2.length > 0)
    {
        while (q1.length > 0)
        {
             
            // Pushing left child of current node
            // in first queue into second queue
            if (q1[0].left != null)
            {
                q2.push(q1[0].left);
            }
             
            // Pushing right child of current node
            // in first queue into second queue
            if (q1[0].right != null)
            {
                q2.push(q1[0].right);
            }
            document.write(q1[0].data + " ");
            q1.shift();
        }
        document.write("<br>");
        while (q2.length > 0)
        {
             
            // Pushing left child of current node
            // in second queue into first queue
            if (q2[0].left != null)
            {
                q1.push(q2[0].left);
            }
             
            // Pushing right child of current
            // node in second queue into first queue
            if (q2[0].right != null)
            {
                q1.push(q2[0].right);
            }
            document.write(q2[0].data + " ");
            q2.shift();
        }
        document.write("<br>");
    }
}
 
// Driver Code
var root = new Node(1);
root.left = new Node(2);
root.right = new Node(3);
root.left.left = new Node(4);
root.left.right = new Node(5);
root.right.right = new Node(6);
 
levelOrder(root);
 
// This code is contributed by noob2000
 

输出 ：

1
2 3
4 5 6

时间复杂度： O(n)