在Java中使用堆栈实现中序、前序、后序
两种类型的数据结构线性数据结构和第二种是非线性数据结构,这些数据结构之间的主要区别在于横向这些数据结构的元素的方式。线性数据结构在一次运行中遍历所有元素,而非线性数据结构以层次结构方式遍历所有元素。
在本文中,我们将使用堆栈数据结构在非线性数据结构的二叉树中实现不同类型的深度优先遍历。
输入二叉树:
深度优先遍历:
- 中序(左、根、右):4 2 5 1 3
- 预购(根、左、右):1 2 4 5 3
- 后序(左、右、根):4 5 2 3 1
A. 中序遍历
1.创建一个空栈 S。
2.以 root 身份初始化当前节点。
3.将当前节点push到S,设置current = current->left,直到current为NULL。
4.如果当前为NULL且栈不为空则
- 从堆栈中弹出顶部项目。
- 打印弹出的item,设置current = popped_item->right
- 转到步骤 3。
5.如果当前为 NULL 并且堆栈为空,那么我们就完成了。
下面是上述方法的实现:
Java
// Non-Recursive Java program for inorder traversal
import java.util.Stack;
// Class containing left and right child of
// current node and key value
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// Class to print the inorder traversal
class BinaryTree {
Node root;
void inorder()
{
if (root == null)
return;
Stack s = new Stack();
Node curr = root;
// traverse the tree
while (curr != null || s.size() > 0) {
// Reach the left most
// Node of the current Node
while (curr != null) {
// place pointer to a tree node on
// the stack before traversing
// the node's left subtree
s.push(curr);
curr = curr.left;
}
// Current must be NULL at this point
curr = s.pop();
System.out.print(curr.data + " ");
// we have visited the node and its
// left subtree. Now, it's right
// subtree's turn
curr = curr.right;
}
}
public static void main(String args[])
{
// Creating a binary tree and
// entering the nodes
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(4);
tree.root.left.right = new Node(5);
tree.inorder();
}
}
Java
// Java Program for Pre order Traversal
import java.util.*;
import java.io.*;
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// Class to print the preorder traversal
class BinaryTree {
Node root;
void preorder()
{
if (root == null)
return;
Stack s = new Stack();
Node curr = root;
// traverse the tree
while (curr != null || s.size() > 0) {
// Reach the left most
// Node of the curr Node
while (curr != null) {
// place pointer to a tree node on
// the stack before traversing
// the node's left subtree
s.push(curr);
// print the peak element
System.out.print(s.peek().data + " ");
curr = curr.left;
}
// Current must be NULL at this point
curr = s.pop();
// we have visited the node and its
// left subtree. Now, it's right
// subtree's turn
curr = curr.right;
}
}
public static void main(String args[])
{
// creating a binary tree and
// entering the nodes
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(4);
tree.root.left.right = new Node(5);
tree.preorder();
}
}
Java
// Non-Recursive Java Program for preorder traversal
import java.util.Stack;
// Class containing left and right child of
// current node and key value
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// Class to print the preorder traversal
class BinaryTree {
Node root;
void preorder()
{
if (root == null) {
return;
}
Stack S = new Stack<>();
// Push root element in the stack
S.add(root);
// print the root element
System.out.print(root.data + " ");
// Push right subtree to the stack
if (root.right != null) {
S.add(root.right);
}
// Push left subtree to the stack
if (root.left != null) {
S.add(root.left);
}
// Iterate till Size of the Stack not equal to 1
while (S.size() != 1) {
// Peek element of the stack
Node temp = S.peek();
// pop the element from the stack
S.pop();
if (temp != null) {
// print the pop element
System.out.print(temp.data + " ");
// Push right subtree of the pop element
if (temp.right != null) {
S.add(temp.right);
}
// Push left subtree of the pop element
if (temp.left != null) {
S.add(temp.left);
}
}
}
}
public static void main(String args[])
{
// creating a binary tree and
// entering the nodes
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(4);
tree.root.left.right = new Node(5);
tree.preorder();
}
}
Java
// Java program for iterative postorder
// traversal using stack
import java.util.ArrayList;
import java.util.Stack;
// A binary tree node
class Node {
int data;
Node left, right;
Node(int item)
{
data = item;
left = right;
}
}
class BinaryTree {
Node root;
// An iterative function to do postorder traversal
// of a given binary tree
void postOrder(Node node)
{
Stack S = new Stack();
// Check for empty tree
if (node == null)
return;
S.push(node);
Node prev = null;
while (!S.isEmpty()) {
Node current = S.peek();
// go down the tree in search of a leaf an if so
// process it and pop stack otherwise move down
if (prev == null || prev.left == current
|| prev.right == current) {
if (current.left != null)
S.push(current.left);
else if (current.right != null)
S.push(current.right);
else {
S.pop();
System.out.print(current.data + " ");
}
// go up the tree from left node, if the
// child is right push it onto stack
// otherwise process parent and pop
}
else if (current.left == prev) {
if (current.right != null)
S.push(current.right);
else {
S.pop();
System.out.print(current.data + " ");
}
// go up the tree from right node and after
// coming back from right node process parent
// and pop stack
}
else if (current.right == prev) {
S.pop();
System.out.print(current.data + " ");
}
prev = current;
}
}
// Driver program to test above functions
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
// Let us create trees shown in above diagram
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
tree.postOrder(tree.root);
}
}
4 2 5 1 3
- 时间复杂度: O(n)
- 空间复杂度: O(n)
B. 前序遍历
使用 Stack 数据结构对二叉树进行预序遍历有两种方法:
方法1:这种方法与上面Transversal中讨论的完全相同。
- 创建一个空栈 S。
- 将当前节点初始化为 root。
- 将当前节点推到 S 并设置 current = current->left 打印堆栈中的 peek 元素,直到 current 为 NULL。
- 如果 current 为 NULL 且堆栈不为空,则
a) 从堆栈中弹出顶部项目。
b) 设置 current = popped_item->right。
c) 转到步骤 3。 - 如果当前为 NULL 并且堆栈为空,那么我们就完成了。
下面是上述方法的实现:
Java
// Java Program for Pre order Traversal
import java.util.*;
import java.io.*;
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// Class to print the preorder traversal
class BinaryTree {
Node root;
void preorder()
{
if (root == null)
return;
Stack s = new Stack();
Node curr = root;
// traverse the tree
while (curr != null || s.size() > 0) {
// Reach the left most
// Node of the curr Node
while (curr != null) {
// place pointer to a tree node on
// the stack before traversing
// the node's left subtree
s.push(curr);
// print the peak element
System.out.print(s.peek().data + " ");
curr = curr.left;
}
// Current must be NULL at this point
curr = s.pop();
// we have visited the node and its
// left subtree. Now, it's right
// subtree's turn
curr = curr.right;
}
}
public static void main(String args[])
{
// creating a binary tree and
// entering the nodes
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(4);
tree.root.left.right = new Node(5);
tree.preorder();
}
}
1 2 4 5 3
- 时间复杂度: O(n)
- 空间复杂度: O(n)
方法2:在Preorder Transversal中,先打印根元素,然后是左子树,然后是右子树。我们知道Stack数据结构遵循LIFO(Last in First Out),所以我们利用这个栈的这个特性,我们先把当前树的右侧部分压入当前树的左侧部分之后,并且在每次迭代中,我们从堆栈中弹出 peak 元素并打印,然后再次推送 pop 元素的右侧部分和 pop 元素的左侧部分,直到堆栈的大小不等于 1,因为我们已经打印了第一个元素。
算法:
- 创建一个空栈 S。
- 打印根元素。
- 将右子树推入堆栈。
- 将左子树压入堆栈。
- 如果堆栈大小不为 1,则
- 从堆栈中弹出顶部项目。
- 打印元素
- 将弹出元素的右子树推入堆栈。
- 将弹出元素的左子树推入堆栈。
- 如果堆栈的大小为 1,则返回 main 方法
下面是上述方法的实现:
Java
// Non-Recursive Java Program for preorder traversal
import java.util.Stack;
// Class containing left and right child of
// current node and key value
class Node {
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
// Class to print the preorder traversal
class BinaryTree {
Node root;
void preorder()
{
if (root == null) {
return;
}
Stack S = new Stack<>();
// Push root element in the stack
S.add(root);
// print the root element
System.out.print(root.data + " ");
// Push right subtree to the stack
if (root.right != null) {
S.add(root.right);
}
// Push left subtree to the stack
if (root.left != null) {
S.add(root.left);
}
// Iterate till Size of the Stack not equal to 1
while (S.size() != 1) {
// Peek element of the stack
Node temp = S.peek();
// pop the element from the stack
S.pop();
if (temp != null) {
// print the pop element
System.out.print(temp.data + " ");
// Push right subtree of the pop element
if (temp.right != null) {
S.add(temp.right);
}
// Push left subtree of the pop element
if (temp.left != null) {
S.add(temp.left);
}
}
}
}
public static void main(String args[])
{
// creating a binary tree and
// entering the nodes
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(4);
tree.root.left.right = new Node(5);
tree.preorder();
}
}
1 2 4 5 3
- 时间复杂度: O(n)
- 空间复杂度: O(n)
C. 后序遍历
1.创建一个空栈
2.在root不为NULL的情况下执行以下操作
- 将根的右孩子推入堆栈,然后将根推入堆栈。
- 将 root 设置为 root 的左孩子。
3.从堆栈中弹出一个项目并将其设置为根。
- 如果弹出的项目有一个右孩子并且右孩子在堆栈顶部,则将右孩子从堆栈中移除,将根推回,并将根设置为根的右孩子。
- 否则打印 root 的数据并将 root 设置为 NULL。
4.当堆栈不为空时,重复步骤 2 和 3。
下面是上述方法的实现:
Java
// Java program for iterative postorder
// traversal using stack
import java.util.ArrayList;
import java.util.Stack;
// A binary tree node
class Node {
int data;
Node left, right;
Node(int item)
{
data = item;
left = right;
}
}
class BinaryTree {
Node root;
// An iterative function to do postorder traversal
// of a given binary tree
void postOrder(Node node)
{
Stack S = new Stack();
// Check for empty tree
if (node == null)
return;
S.push(node);
Node prev = null;
while (!S.isEmpty()) {
Node current = S.peek();
// go down the tree in search of a leaf an if so
// process it and pop stack otherwise move down
if (prev == null || prev.left == current
|| prev.right == current) {
if (current.left != null)
S.push(current.left);
else if (current.right != null)
S.push(current.right);
else {
S.pop();
System.out.print(current.data + " ");
}
// go up the tree from left node, if the
// child is right push it onto stack
// otherwise process parent and pop
}
else if (current.left == prev) {
if (current.right != null)
S.push(current.right);
else {
S.pop();
System.out.print(current.data + " ");
}
// go up the tree from right node and after
// coming back from right node process parent
// and pop stack
}
else if (current.right == prev) {
S.pop();
System.out.print(current.data + " ");
}
prev = current;
}
}
// Driver program to test above functions
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
// Let us create trees shown in above diagram
tree.root = new Node(1);
tree.root.left = new Node(2);
tree.root.right = new Node(3);
tree.root.left.left = new Node(4);
tree.root.left.right = new Node(5);
tree.postOrder(tree.root);
}
}
4 5 2 3 1
- 时间复杂度: O(n)
- 空间复杂度: O(n)