二叉树的逆时针螺旋遍历

给定一棵二叉树。任务是打印给定二叉树的循环逆时针螺旋顺序遍历。
逆时针遍历是指从树的底部而不是顶部的根节点以顺时针方向螺旋地遍历树。
例子：

Input : 
          1
         /  \
        2    3
       / \    \
      4   5    6
         /    / \
        7    8   9
Output : 9 8 7 1 6 5 4 2 3

Input :
           20
         /   \
        8     22
      /   \  /   \
     5     3 4    25
          / \
         10  14
Output : 14 10 20 25 4 3 5 8 22

方法：这个想法是使用两个变量i初始化为 1 和j初始化为树的高度，并运行一个 while 循环，直到i 变得大于 j才会中断。

• 使用另一个变量flag并将其初始化为 0。现在在 while 循环中，检查一个条件，如果flag等于 0，则从右到左遍历树并将flag 标记为 1，以便下次我们从左到对，这样做是为了保持螺旋顺时针遍历顺序。
• 然后减少 j 的值，以便下次访问刚好高于当前级别的级别。
• 此外，当我们从顶部遍历该级别时，我们会将flag 标记为 0，以便下次我们从右到左遍历树，然后增加 i 的值，以便下次我们访问低于当前级别的级别。
• 重复整个过程，直到完全遍历完二叉树。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
 
// Binary tree node
struct Node {
    struct Node* left;
    struct Node* right;
    int data;
 
    Node(int data)
    {
        this->data = data;
        this->left = NULL;
        this->right = NULL;
    }
};
 
// Recursive Function to find height
// of binary tree
int height(struct Node* root)
{
    // Base condition
    if (root == NULL)
        return 0;
 
    // Compute the height of each subtree
    int lheight = height(root->left);
    int rheight = height(root->right);
 
    // Return the maximum of two
    return max(1 + lheight, 1 + rheight);
}
 
// Function to Print Nodes from left to right
void leftToRight(struct Node* root, int level)
{
    if (root == NULL)
        return;
 
    if (level == 1)
        cout << root->data << " ";
 
    else if (level > 1) {
        leftToRight(root->left, level - 1);
        leftToRight(root->right, level - 1);
    }
}
 
// Function to Print Nodes from right to left
void rightToLeft(struct Node* root, int level)
{
    if (root == NULL)
        return;
 
    if (level == 1)
        cout << root->data << " ";
 
    else if (level > 1) {
        rightToLeft(root->right, level - 1);
        rightToLeft(root->left, level - 1);
    }
}
 
// Function to print reverse clockwise spiral
// traversal of a binary tree
void ReverseClockWiseSpiral(struct Node* root)
{
    int i = 1;
    int j = height(root);
 
    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 0;
    while (i <= j) {
 
        // If flag is zero print nodes
        // from right to left
        if (flag == 0) {
            rightToLeft(root, j);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 1;
 
            // Decrement j
            j--;
        }
 
        // If flag is one print nodes
        // from left to right
        else {
            leftToRight(root, i);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 0;
 
            // Increment i
            i++;
        }
    }
}
 
// Driver code
int main()
{
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->right->left = new Node(6);
    root->right->right = new Node(7);
    root->left->right = new Node(5);
 
    ReverseClockWiseSpiral(root);
 
    return 0;
}

Java

// Java implementation of the approach
import java.util.*;
 
class GFG
{
     
// Binary tree node
static class Node
{
    Node left;
    Node right;
    int data;
 
    Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
// Recursive Function to find height
// of binary tree
static int height( Node root)
{
    // Base condition
    if (root == null)
        return 0;
 
    // Compute the height of each subtree
    int lheight = height(root.left);
    int rheight = height(root.right);
 
    // Return the maximum of two
    return Math.max(1 + lheight, 1 + rheight);
}
 
// Function to Print Nodes from left to right
static void leftToRight( Node root, int level)
{
    if (root == null)
        return;
 
    if (level == 1)
        System.out.print( root.data + " ");
 
    else if (level > 1)
    {
        leftToRight(root.left, level - 1);
        leftToRight(root.right, level - 1);
    }
}
 
// Function to Print Nodes from right to left
static void rightToLeft( Node root, int level)
{
    if (root == null)
        return;
 
    if (level == 1)
        System.out.print( root.data + " ");
 
    else if (level > 1)
    {
        rightToLeft(root.right, level - 1);
        rightToLeft(root.left, level - 1);
    }
}
 
// Function to print reverse clockwise spiral
// traversal of a binary tree
static void ReverseClockWiseSpiral( Node root)
{
    int i = 1;
    int j = height(root);
 
    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 0;
    while (i <= j)
    {
 
        // If flag is zero print nodes
        // from right to left
        if (flag == 0)
        {
            rightToLeft(root, j);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 1;
 
            // Decrement j
            j--;
        }
 
        // If flag is one print nodes
        // from left to right
        else
        {
            leftToRight(root, i);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 0;
 
            // Increment i
            i++;
        }
    }
}
 
// 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.right.left = new Node(6);
    root.right.right = new Node(7);
    root.left.right = new Node(5);
 
    ReverseClockWiseSpiral(root);
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 implementation of the approach
  
# Binary tree node
class Node:
     
    def __init__(self,data):
         
        self.left = None
        self.right = None
        self.data = data;
         
# Recursive Function to find height
# of binary tree
def height(root):
 
    # Base condition
    if (root == None):
        return 0;
  
    # Compute the height of each subtree
    lheight = height(root.left);
    rheight = height(root.right);
  
    # Return the maximum of two
    return max(1 + lheight, 1 + rheight);
 
# Function to Print Nodes from
# left to right
def leftToRight(root, level):
 
    if (root == None):
        return;
  
    if (level == 1):
        print(root.data, end = " ")
  
    elif (level > 1):
        leftToRight(root.left, level - 1);
        leftToRight(root.right, level - 1);
         
# Function to Print Nodes
# from right to left
def rightToLeft(root, level):
 
    if (root == None):
        return;
  
    if (level == 1):
        print(root.data, end = " ")
  
    elif (level > 1):
        rightToLeft(root.right, level - 1);
        rightToLeft(root.left, level - 1);
         
# Function to print Reverse clockwise
# spiral traversal of a binary tree
def ReverseClockWiseSpiral(root):
 
    i = 1;
    j = height(root);
  
    # Flag to mark a change in the
    # direction of printing nodes
    flag = 0;
     
    while (i <= j):
  
        # If flag is zero print nodes
        # from right to left
        if (flag == 0):
            rightToLeft(root, j);
  
            # Set the value of flag as zero
            # so that nodes are next time
            # printed from left to right
            flag = 1;
  
            # Increment i
            j -= 1;
         
        # If flag is one print nodes
        # from left to right
        else:
            leftToRight(root, i);
  
            # Set the value of flag as zero
            # so that nodes are next time
            # printed from right to left
            flag = 0;
  
            # Decrement j
            i += 1;
 
# Driver code
if __name__=="__main__":
     
    root = Node(1);
    root.left = Node(2);
    root.right = Node(3);
    root.left.left = Node(4);
    root.right.left = Node(6);
    root.right.right = Node(7);
    root.left.right = Node(5);
  
    ReverseClockWiseSpiral(root);
 
# This code is contributed by rutvik_56

C#

// C# implementation of the approach
using System;
 
class GFG
{
     
// Binary tree node
public class Node
{
    public Node left;
    public Node right;
    public int data;
 
    public Node(int data)
    {
        this.data = data;
        this.left = null;
        this.right = null;
    }
};
 
// Recursive Function to find height
// of binary tree
static int height( Node root)
{
    // Base condition
    if (root == null)
        return 0;
 
    // Compute the height of each subtree
    int lheight = height(root.left);
    int rheight = height(root.right);
 
    // Return the maximum of two
    return Math.Max(1 + lheight, 1 + rheight);
}
 
// Function to Print Nodes from left to right
static void leftToRight( Node root, int level)
{
    if (root == null)
        return;
 
    if (level == 1)
        Console.Write( root.data + " ");
 
    else if (level > 1)
    {
        leftToRight(root.left, level - 1);
        leftToRight(root.right, level - 1);
    }
}
 
// Function to Print Nodes from right to left
static void rightToLeft( Node root, int level)
{
    if (root == null)
        return;
 
    if (level == 1)
        Console.Write( root.data + " ");
 
    else if (level > 1)
    {
        rightToLeft(root.right, level - 1);
        rightToLeft(root.left, level - 1);
    }
}
 
// Function to print reverse clockwise spiral
// traversal of a binary tree
static void ReverseClockWiseSpiral( Node root)
{
    int i = 1;
    int j = height(root);
 
    // Flag to mark a change in the direction
    // of printing nodes
    int flag = 0;
    while (i <= j)
    {
 
        // If flag is zero print nodes
        // from right to left
        if (flag == 0)
        {
            rightToLeft(root, j);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from left to right
            flag = 1;
 
            // Decrement j
            j--;
        }
 
        // If flag is one print nodes
        // from left to right
        else
        {
            leftToRight(root, i);
 
            // Set the value of flag as zero
            // so that nodes are next time
            // printed from right to left
            flag = 0;
 
            // Increment i
            i++;
        }
    }
}
 
// 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.right.left = new Node(6);
    root.right.right = new Node(7);
    root.left.right = new Node(5);
 
    ReverseClockWiseSpiral(root);
}
}
 
// This code contributed by Rajput-Ji

Javascript

 
    // JavaScript implementation of the approach
     
    // Binary tree node
    class Node
    {
        constructor(data) {
           this.left = null;
           this.right = null;
           this.data = data;
        }
    }
     
    // Recursive Function to find height
    // of binary tree
    function height(root)
    {
        // Base condition
        if (root == null)
            return 0;
 
        // Compute the height of each subtree
        let lheight = height(root.left);
        let rheight = height(root.right);
 
        // Return the maximum of two
        return Math.max(1 + lheight, 1 + rheight);
    }
 
    // Function to Print Nodes from left to right
    function leftToRight(root, level)
    {
        if (root == null)
            return;
 
        if (level == 1)
            document.write( root.data + " ");
 
        else if (level > 1)
        {
            leftToRight(root.left, level - 1);
            leftToRight(root.right, level - 1);
        }
    }
 
    // Function to Print Nodes from right to left
    function rightToLeft(root, level)
    {
        if (root == null)
            return;
 
        if (level == 1)
            document.write( root.data + " ");
 
        else if (level > 1)
        {
            rightToLeft(root.right, level - 1);
            rightToLeft(root.left, level - 1);
        }
    }
 
    // Function to print reverse clockwise spiral
    // traversal of a binary tree
    function ReverseClockWiseSpiral(root)
    {
        let i = 1;
        let j = height(root);
 
        // Flag to mark a change in the direction
        // of printing nodes
        let flag = 0;
        while (i <= j)
        {
 
            // If flag is zero print nodes
            // from right to left
            if (flag == 0)
            {
                rightToLeft(root, j);
 
                // Set the value of flag as zero
                // so that nodes are next time
                // printed from left to right
                flag = 1;
 
                // Decrement j
                j--;
            }
 
            // If flag is one print nodes
            // from left to right
            else
            {
                leftToRight(root, i);
 
                // Set the value of flag as zero
                // so that nodes are next time
                // printed from right to left
                flag = 0;
 
                // Increment i
                i++;
            }
        }
    }
     
    let root = new Node(1);
    root.left = new Node(2);
    root.right = new Node(3);
    root.left.left = new Node(4);
    root.right.left = new Node(6);
    root.right.right = new Node(7);
    root.left.right = new Node(5);
  
    ReverseClockWiseSpiral(root);
 

输出：

7 6 5 4 1 3 2    

时间复杂度：O(N^2)，其中 N 是二叉树中节点的总数。
辅助空间： O(N)