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。