螺旋形式的层次顺序遍历

螺旋形式的层次顺序遍历

什么是二叉树?

二叉树是一种由层次化组织的节点组成的数据结构。每个节点最多有两个子节点,通常是左子和右子。根节点是树中最高的节点,叶节点是树底部没有子节点的节点。

下面是一个二叉树的简单例子:

    1
   / \
  2   3
 / \ / \
4  5 6  7

该二叉树有7个节点,节点1是根节点,节点4、5、6和7是叶节点。

下面是一个二叉树的伪代码实现:

// Define a Node class representing a binary tree node.
class Node {
  // Define properties for the node value left child and right child.
  public int value;
  public Node left;
  public Node right;

  // Define a constructor that initializes the node value and children.
  public Node(int value, Node left, Node right) {
    this.value = value;
    this.left = left;
    this.right = right;
  }
}

// Define a BinaryTree class that represents a binary tree.
class BinaryTree {
  // Define a property for the root node of the tree.
  public Node root;

  // Define a constructor that initializes the root node.
  public BinaryTree(Node root) {
    this.root = root;
  }
}

什么是层次顺序遍历?

二叉树的层序遍历涉及按特定顺序访问树的节点。在普通的层序遍历中,我们从左到右,从根节点开始,向下移动,访问每一层的节点。在普通的层序遍历中,我们会按照以下顺序访问节点:1、2、3、4、5、6、7。这被称为层次顺序遍历,因为我们在进入下一层之前会访问树的每一层的节点。

要执行分层遍历,我们可以使用一个队列来存储每层的节点。我们首先将根节点推入队列,然后从左到右遍历当前层的节点。在访问完当前层的所有节点后,我们再进入下一层。

螺旋形式的层次顺序遍历是树遍历的一种,它以螺旋模式访问树中的每个节点。这种类型的遍历对于寻找两个节点之间的最短路径非常有用,因为它允许算法探索所有可能的路径,而不需要回溯。螺旋模式的遍历对于可视化树的结构也很有用,因为它可以让用户更清楚地看到节点之间的关系。螺旋形式的层次顺序遍历从根节点开始,然后以螺旋模式访问每个节点。算法从访问根节点开始,然后顺时针绕树移动,按顺序访问每个节点。在访问完每个节点后,算法移动到树的下一层,重复这个过程,直到访问完所有节点。螺旋形式的层级顺序遍历是一种递归算法,这意味着它反复调用自己,直到所有节点都被访问过。

什么是螺旋形?

在二叉树中,“螺旋形”遍历树,其中每一层的节点都以螺旋模式访问。这意味着对于树的每一层,节点的访问顺序是从左到右,然后从右到左,再从左到右,以此类推。

螺旋形二叉树结构

    1
   / \
  2   3
 / \ / \
4  5 6  7

解释

螺旋形式的树的层次-顺序遍历将按以下顺序访问节点:1、2、3、4、5、6、7。

这与普通的层次遍历不同,后者会以1、2、3、4、5、6、7的顺序访问节点。

要以螺旋形式对二叉树进行逐层遍历,可以使用队列和栈。你可以将根节点添加到队列中,然后执行以下步骤,直到队列为空:

从队列中取出顶部元素,并将其添加到结果列表中。

如果节点有左子节点,则将其添加到队列中。

如果节点有正确的子节点,则将其添加到堆栈中。

如果队列为空,而栈不是,则弹出栈中的所有元素,并将它们添加到队列中。

该算法允许您以螺旋模式访问每个层次上的节点,如上所述。

伪代码

二叉树的层序遍历涉及按特定顺序访问树的节点。在普通的层次顺序遍历中,我们从左到右访问每一层的节点。在螺旋形式的层次顺序遍历中,我们以交替的方向(从左到右,然后从右到左,以此类推)访问每个层次的节点。

// Initialize the queue with the root node
queue.enqueue(root)
// Initialize the direction of traversal to left-to-right
direction = 1
// Loop until the queue is empty
while (the queue is not empty)
  // Calculate the number of nodes in the current level
  count = queue.size()

  // Traverse all the nodes in the current level
  for (i = 1 to count)
    // Dequeue a node from the queue
    node = queue.dequeue()

    // Print the node data
    print(node.data)

    // Enqueue the left and right children of the node, if they exist
    if (node.left) queue.enqueue(node.left)
    if (node.right) queue.enqueue(node.right)

  // Reverse the direction of traversal
  direction = -direction

C++ 代码

// Here we are writing down the C++ programming language code to demonstrate
// the concept of Level order traversal in spiral form with its relevant 
// code and output supported by syntax 
#include 
using namespace std;
// defining the structure of the node from the below code
struct node {
    int data;
    // assigning the initial nodes to the left and right
    struct node* left;
           // structure node allocated dynamically to right
    struct node* right;
};
// below, we are writing down the void function to return us the value of the root node, level 
// structurally
void printGivenLevel(struct node* root, int level, int ltr);
// defining the height calling recursively with structure node as the parameter
int height(struct node* node);
// structure node dynamics
struct node* newNode(int data);
void printSpiral(struct node* root)
{
    // defining the height for the structure node
    int h = height(root);
    int i;
    bool ltr = false;
// running the for loop to print the level
    for (i = 1; i <= h; i++) {
        printGivenLevel(root, i, ltr);
        ltr = !ltr;
    }
}
// defining the void function to help us with defining the struct, and level from here
void printGivenLevel(struct node* root, int level, int ltr)
{
// conditional solving edge case if the root is null
    if (root == NULL)
        return;
// conditional solving if the level becomes one
    if (level == 1)
        cout << root->data << " ";
// edge case condition if level comes greater than one
    else if (level > 1) {
        if (ltr) {
            printGivenLevel(root->left, level - 1, ltr);
            printGivenLevel(root->right, level - 1, ltr);
        }
        else {
            printGivenLevel(root->right, level - 1, ltr);
            printGivenLevel(root->left, level - 1, ltr);
        }
    }
}
// defining the height function with a return type integer
int height(struct node* node)
{
    if (node == NULL)
        return 0;
// edge case condition if the level becomes greater than one
    else {
        int lheight = height(node->left);
        int rheight = height(node->right);
        if (lheight > rheight)
            return (lheight + 1);
        else
            return (rheight + 1);
    }
}
struct node* newNode(int data)
{
    node* newnode = new node();
    newnode->data = data;
    newnode->left = NULL;
    newnode->right = NULL;

    return (newnode);
}
// The main driver code functionality starts from here
int main()
{
    struct node* root = newNode(1);
    // defining the allocation of the root nodes
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(7);
    root->left->right = newNode(6);
    root->right->left = newNode(5);
    root->right->right = newNode(4);
    // printing the spiral nodes traversal
    printf("Spiral Order traversal of "
        "binary tree is \n");

    printSpiral(root);

    return 0;
// The main driver code functionality ends from here    

}

输出:

Spiral Order traversal of binary tree is 
1 2 3 4 5 6 7

C 代码

// Here we are writing down the C programming language code to demonstrate
// the concept of Level order traversal in spiral form with its relevant 
// code and output supported by syntax 
#include 
#include 
#include 
// defining the structure of the node from the below code
struct node {
    int data;
    // assigning the initial nodes to the left and right
    struct node* left;
    struct node* right;
};
void printGivenLevel(struct node* root, int level, int ltr);
int height(struct node* node);
// structure node dynamics
struct node* newNode(int data);
void printSpiral(struct node* root)
{

    // defining the height for the structure node
    int h = height(root);
    int i;
    bool ltr = false;
    // running the for loop to print the level
    for (i = 1; i <= h; i++) {
        printGivenLevel(root, i, ltr);
        ltr = !ltr;
    }
}
// defining the void function to help us with defining the struct, level
void printGivenLevel(struct node* root, int level, int ltr)
{
    if (root == NULL)
        return;
    if (level == 1)
        printf("%d ", root->data);
        // edge case condition if the level becomes greater than one
    else if (level > 1) {
        if (ltr) {
            printGivenLevel(root->left, level - 1, ltr);
            printGivenLevel(root->right, level - 1, ltr);
        }
        else {
            printGivenLevel(root->right, level - 1, ltr);
            printGivenLevel(root->left, level - 1, ltr);
        }
    }
}
// defining the height function with a return type integer
int height(struct node* node)
{
    if (node == NULL)
        return 0;
        // edge case condition if the level becomes greater than one
    else {
        int lheight = height(node->left);
        int rheight = height(node->right);
        if (lheight > rheight)
            return (lheight + 1);
        else
            return (rheight + 1);
    }
}
struct node* newNode(int data)
{
    struct node* node
        = (struct node*)malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;

    return (node);
}
// The main driver code functionality starts from here
int main()
{
    struct node* root = newNode(1);
    // defining the allocation of the root nodes
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(7);
    root->left->right = newNode(6);
    root->right->left = newNode(5);
    root->right->right = newNode(4);
    // printing the spiral nodes traversal
    printf("Spiral Order traversal of binary tree is \n");
    printSpiral(root);

    return 0;
// The main driver code functionality ends from here
}

输出:

Spiral Order traversal of binary tree is 
1 2 3 4 5 6 7

该算法从访问根节点开始,然后顺时针绕树移动,依次访问每个节点。在访问完每个节点后,算法移动到树的下一层,重复这个过程,直到访问完所有节点。螺旋形式的层序遍历是一种高效的算法,因为它只访问树中的每个节点一次。这使得它非常适合树结构庞大而复杂的应用程序,允许算法探索所有可能的路径,而无需回溯。

C# 代码

// Perform a level order traversal of the given binary tree in spiral form.
// Return a list containing the traversed nodes.
public List SpiralLevelOrderTraversal(TreeNode root)
{
    // Initialize the result list and the queue and stack.
    List result = new List();
    Queue queue = new Queue();
    Stack stack = new Stack();

    // Add the root node to the queue.
    queue.Enqueue(root);

    // Perform the spiral traversal.
    while (queue.Count > 0 || stack.Count > 0)
    {
        // Take the top element from the queue and add it to the result list.
        TreeNode current = queue.Dequeue();
        result.Add(current.val);

        // If the node has a left child, add it to the queue.
        if (current.left != null)
        {
            queue.Enqueue(current.left);
        }

        // If the node has a right child, add it to the stack.
        if (current.right != null)
        {
            stack.Push(current.right);
        }

        // If the queue is empty and the stack is not, pop all elements from the stack
        // and add them to the queue.
        if (queue.Count == 0 && stack.Count > 0)
        {
            while (stack.Count > 0)
            {
                queue.Enqueue(stack.Pop());
            }
        }
    }

    // Return the result list.
    return result;
}

c#中的算法解释

注意,这个实现假设你有一个表示二叉树中节点的TreeNode类,其中val、left和right属性分别存储该节点的值,并引用该节点的左子节点和右子节点。这个伪代码定义了一个表示二叉树中节点的节点类,具有节点值、左子节点和右子节点的属性。它还定义了一个表示binary are的BinaryTree类,具有树的节点的属性。

要使用这个实现,你可以创建一个BinaryTree对象,并向构造函数传递一个表示根节点的Node对象,从而创建一棵二叉树。然后,你可以通过设置父节点的left和right属性来引用左、右子节点,从而向树中添加节点。

Java 代码

// Define a node class to represent a node in the tree.
class Node {
    int data;
    Node left, right;

    public Node(int d)
    {
        data = d;
        left = right = null;
    }
}
// Define a class to represent the tree
class BinaryTree {
    Node root;

    // This method performs a level order traversal in spiral form.
    // It starts at the root node and prints the data in each node,
    // in spiral order.
    void printSpiral(Node node)
    {
        int h = height(node);
        int i;

        boolean ltr = false;
        for (i = 1; i <= h; i++) {
            printGivenLevel(node, i, ltr);


            ltr = !ltr;
        }
    }
// If the current level stack is empty, it means we have
            //I finished visiting all the nodes on the current level, so
            // we need to switch to the next level.

    int height(Node node)
    {
        if (node == null)
            return 0;
        else {


            int lheight = height(node.left);
            int rheight = height(node.right);


            if (lheight > rheight)
                return (lheight + 1);
            else
                return (rheight + 1);
        }
    }


    void printGivenLevel(Node node, int level, boolean ltr)
    {
        // We will use two stacks to keep track of the nodes to visit.
        // One stack will hold the nodes on the current level, and the
        // other will hold the nodes on the next level.
        if (node == null)
            return;
        if (level == 1)
            System.out.print(node.data + " ");
        else if (level > 1) {
             // Add the node's children to the next level stack in the
            // appropriate order depending on whether we are currently
            // traversing the current level from left to right or from
            // right to left.
            if (ltr != false) {
                printGivenLevel(node.left, level - 1, ltr);
                printGivenLevel(node.right, level - 1, ltr);
            }
            else {
                printGivenLevel(node.right, level - 1, ltr);
                printGivenLevel(node.left, level - 1, ltr);
            }
        }
    }

    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root. left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(7);
        tree.root.left.right = new Node(6);
        tree.root.right.left = new Node(5);
        tree.root.right.right = new Node(4);
        System.out.println(
            "Spiral order traversal of Binary Tree is ");
        tree.printSpiral(tree.root);
    }
}

输出:

The spiral order traversal of Binary Tree is 1 2 3 4 5 6 7

此外,螺旋模式的遍历对于可视化树的结构很有用,因为它允许用户更清楚地看到节点之间的关系。综上所述,螺旋形式的层次顺序遍历是树遍历的一种类型,它以螺旋模式访问树中的每个节点。这种类型的遍历对于寻找两个节点之间的最短路径非常有用,因为它允许算法在不回溯的情况下探索所有可能的路径。此外,螺旋模式的遍历对于可视化树的结构很有用,因为它允许用户更清楚地看到节点之间的关系。

JavaScript 代码

function levelOrderSpiral(root) {
  if (!root) return;

  // Create an empty queue for level order traversal
  let queue = [];

  // Push the root node
  queue.push(root);

  // Track the direction of traversal. We start from left to right
  let direction = 1;

  // Run while the queue is not empty
  while (queue.length > 0) {
    // Calculate the number of nodes in current level
    let count = queue.length;

    // Traverse all nodes of current level
    for (let i = 0; i < count; i++) {
      // Dequeue a node from queue
      let node = queue.shift();

      // Print the node data
      console.log(node.data);

      // Enqueue left child
      if (node.left) queue.push(node.left);

      // Enqueue right child
      if (node.right) queue.push(node.right);
    }

    // Reverse the direction
    direction *= -1;
  }
}

解释:

这里,我们使用一个队列来存储每层的节点,并使用一个可变的direction来跟踪遍历的方向(从左到右或从右到左)。在每一层的末尾,我们反转方向。如果用二叉树的根节点调用这个函数,它会以螺旋级顺序打印树中所有节点的数据。

如果树如下所示,则输出为

    1
   / \
  2   3
 / \ / \
4  5 6  7

输出:

1
3
2
4
5
6
7

javascript中的层次顺序遍历

在普通的层次顺序遍历中,我们会按照以下顺序访问节点:1、2、3、4、5、6、7。

在螺旋形式的层次遍历中,我们将按照以下顺序访问节点:1、3、2、4、5、6、7。

为了以螺旋形式进行逐层遍历,我们可以使用一个队列来存储每层的节点。我们首先将根节点推入队列,然后从左到右(或者从右到左,取决于遍历的方向)遍历当前层的节点。在当前层的所有节点都被访问过之后,我们反转遍历的方向,继续前进到下一层。这个算法的时间复杂度为O(n),其中n是树中的节点数,因为我们对树中的每个节点都只访问一次。由于我们使用队列来存储每一层的节点,它也具有O(n)的空间复杂度。

Python 代码

class newNode:
 # Define a Node class to represent a node in the tree.

    def __init__(self, key):
        self.data = key
        self.left = None
        self.right = None


 # This method performs a level order traversal in spiral form.
    # It starts at the root node and prints the data in each node,
    # in spiral order.


def printSpiral(root):
# Define a Tree class to represent the tree.
    h = height(root)


    ltr = False
    for i in range(1, h + 1):

        printGivenLevel(root, i, ltr)

        ltr = not ltr

# We will use two stacks to keep track of the nodes to visit.
        # One stack will hold the nodes on the current level, and the
        # other will hold the nodes on the next level.



def printGivenLevel(root, level, ltr):

    # Keep track of whether we are traversing the current level from
        # left to right or from right to left. Initially, we start from
        # left to right.

    if(root == None):
        return
    if(level == 1):
        print(root.data, end=" ")
    elif (level > 1):

        if(ltr):
            printGivenLevel(root.left, level - 1, ltr)
            printGivenLevel(root.right, level - 1, ltr)
# Add the node's children to the next level stack, in the
            # appropriate order depending on whether we are currently
            # traversing the current level from left to right or from
            # right to left.
        else:
            printGivenLevel(root.right, level - 1, ltr)
            printGivenLevel(root.left, level - 1, ltr)





def height(node):

    if (node == None):
        return 0
    else:


        lheight = height(node.left)
        rheight = height(node.right)

     # If the current level stack is empty, it means we have
            # finished visiting all the nodes on the current level, so
            # we need to switch to the next level.  
        if (lheight > rheight):
            return(lheight + 1)
        else:
            return(rheight + 1)



if __name__ == '__main__':
    root = newNode(1)
    root.left = newNode(2)
    root.right = newNode(3)
    root.left.left = newNode(7)
    root.left.right = newNode(6)
    root.right.left = newNode(5)
    root.right.right = newNode(4)
    print("Spiral Order traversal of binary tree is")
    printSpiral(root)

输出:

Spiral order traversal of Binary Tree is 1 2 3 4 5 6 7

解释

螺旋形式的层次顺序遍历是一种以层次顺序遍历树节点的方法,但以螺旋模式而不是线性模式。这意味着每一层上的节点都是以圆形的模式访问的,从左到右,然后再回到右再到左,以此类推。要执行螺旋形式的层次顺序遍历,我们可以使用与普通的层次顺序遍历类似的方法。我们从根节点开始,从左到右访问当前层上的节点。然后,对于当前层的每个节点,我们将其子节点添加到队列或堆栈(在本例中我们将使用堆栈),以跟踪下一层的节点。

在访问完当前层的所有节点后,我们切换到下一层,并以相反的方向访问它的节点(从右到左,而不是从左到右)。然后,对于这一层上的每个节点,我们以相反的顺序(先右子,再左子)将其子节点添加到堆栈中。我们以这种方式在层次和方向之间保持交替,直到我们已经访问了树中的所有节点。这将导致螺旋式的遍历模式,每一层的节点都被循环访问。

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