迭代后序遍历 |设置 3
我们已经看到了在二叉树上执行后序遍历的不同方法。
- 后订单遍历。
- 使用两个堆栈的迭代后序遍历。
- 使用一个堆栈的迭代后序遍历。
这是使用单个堆栈迭代地对二叉树执行后序遍历的另一种方法。
考虑以下术语：
0 - Left element
1 - Right element
2 - Node element下面是详细的算法：
Take a Stack and perform the below operations:
1) Insert a pair of the root node as (node, 0).
2) Pop the top element to get the pair
(Let a = node and b be the variable)
If b is equal to 0:
Push another pair as (node, 1) and
Push the left child as (node->left, 0)
Repeat Step 2
Else If b is equal to 1:
Push another pair as (node, 2) and
Push right child of node as (node->right, 0)
Repeat Step 2
Else If b is equal to 2:
Print(node->data)
3) Repeat the above steps while stack is not empty考虑下面只有 3 个节点的二叉树：
插图：
1) Push(a, 0)
Stack - (a, 0)
2) top = (a, 0)
Push(a, 1)
Push(b, 0)
Stack - (b, 0)
(a, 1)
3) top = (b, 0)
Push(b, 1)
Stack - (b, 1)
(a, 1)
4) top = (b, 1)
Push(b, 2)
Stack - (b, 2)
(a, 1)
5) top = (b, 2)
print(b)
Stack -(a, 1)
6) top = (a, 1)
push(a, 2)
push(c, 0)
Stack - (c, 0)
(a, 2)
7) top = (c, 0)
push(c, 1)
Stack - (c, 1)
(a, 2)
8) top = (c, 1)
push(c, 2)
Stack - (c, 2)
(a, 2)
9) top = (c, 2)
print(c)
Stack - (a, 2)
10) top = (a, 2)
print(a)
Stack - empty()下面是上述方法的实现：
C++
// C++ program to print postorder traversal
// iteratively
#include
using namespace std;
// Binary Tree Node
struct Node {
int data;
struct Node* left;
struct Node* right;
};
// Helper function to create a
// new Binary Tree node
struct Node* newNode(int data)
{
struct Node* node = new Node;
node->data = data;
node->left = NULL;
node->right = NULL;
return (node);
}
// Function to perform postorder
// traversal iteratively
void iterativePostorder(Node* root)
{
stack > st;
st.push(make_pair(root, 0));
while (!st.empty()) {
struct Node* temp = st.top().first;
int b = st.top().second;
st.pop();
if (temp == NULL)
continue;
if (b == 0) {
st.push(make_pair(temp, 1));
if (temp->left != NULL)
st.push(make_pair(temp->left, 0));
}
else if (b == 1) {
st.push(make_pair(temp, 2));
if (temp->right != NULL)
st.push(make_pair(temp->right, 0));
}
else
cout << temp->data << " ";
}
}
// Driver Code
int main()
{
// Construct the below Binary Tree
// 1
// / \
// 2 3
// / \
// 4 5
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
iterativePostorder(root);
return 0;
} Java
// Java program to print postorder traversal
// iteratively
import java.util.*;
class GFG
{
//pair class
static class Pair
{
Node first;
int second;
Pair(Node a,int b)
{
first = a;
second = b;
}
}
// Binary Tree Node
static class Node
{
int data;
Node left;
Node right;
};
// Helper function to create a
// new Binary Tree node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Function to perform postorder
// traversal iteratively
static void iterativePostorder(Node root)
{
Stack st = new Stack();
st.add(new Pair(root, 0));
while (st.size() > 0)
{
Node temp = st.peek().first;
int b = st.peek().second;
st.pop();
if (temp == null)
continue;
if (b == 0)
{
st.add(new Pair(temp, 1));
if (temp.left != null)
st.add(new Pair(temp.left, 0));
}
else if (b == 1)
{
st.add(new Pair(temp, 2));
if (temp.right != null)
st.add(new Pair(temp.right, 0));
}
else
System.out.print( temp.data + " ");
}
}
// Driver Code
public static void main(String args[])
{
// Construct the below Binary Tree
// 1
// / \
// 2 3
// / \
// 4 5
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
iterativePostorder(root);
}
}
// This code is contributed by Arnab Kundu Python3
# Python program to print postorder traversal
# iteratively
# pair class
class Pair:
def __init__(self, a, b):
self.first = a
self.second = b
# Binary Tree Node
class Node :
def __init__(self):
self.data = 0
self.left = None
self.right = None
# Helper function to create a
# new Binary Tree node
def newNode(data):
node = Node()
node.data = data
node.left = None
node.right = None
return (node)
# Function to perform postorder
# traversal iteratively
def iterativePostorder( root):
st = []
st.append(Pair(root, 0))
while (len(st)> 0):
temp = st[-1].first
b = st[-1].second
st.pop()
if (temp == None):
continue
if (b == 0) :
st.append(Pair(temp, 1))
if (temp.left != None):
st.append(Pair(temp.left, 0))
elif (b == 1):
st.append(Pair(temp, 2))
if (temp.right != None):
st.append(Pair(temp.right, 0))
else:
print( temp.data ,end= " ")
# Driver Code
# Construct the below Binary Tree
# 1
# / \
# 2 3
# / \
# 4 5
root = newNode(1)
root.left = newNode(2)
root.right = newNode(3)
root.left.left = newNode(4)
root.left.right = newNode(5)
iterativePostorder(root)
# This code is contributed by Arnab KunduC#
// C# program to print postorder traversal
// iteratively
using System;
using System.Collections.Generic;
class GFG
{
//pair class
public class Pair
{
public Node first;
public int second;
public Pair(Node a,int b)
{
first = a;
second = b;
}
}
// Binary Tree Node
public class Node
{
public int data;
public Node left;
public Node right;
};
// Helper function to create a
// new Binary Tree node
static Node newNode(int data)
{
Node node = new Node();
node.data = data;
node.left = null;
node.right = null;
return (node);
}
// Function to perform postorder
// traversal iteratively
static void iterativePostorder(Node root)
{
Stack st = new Stack();
st.Push(new Pair(root, 0));
while (st.Count > 0)
{
Node temp = st.Peek().first;
int b = st.Peek().second;
st.Pop();
if (temp == null)
continue;
if (b == 0)
{
st.Push(new Pair(temp, 1));
if (temp.left != null)
st.Push(new Pair(temp.left, 0));
}
else if (b == 1)
{
st.Push(new Pair(temp, 2));
if (temp.right != null)
st.Push(new Pair(temp.right, 0));
}
else
Console.Write(temp.data + " ");
}
}
// Driver Code
public static void Main(String []args)
{
// Construct the below Binary Tree
// 1
// / \
// 2 3
// / \
// 4 5
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
iterativePostorder(root);
}
}
// This code is contributed by 29AjayKumar Javascript
输出：
4 5 2 3 1时间复杂度： O(N)
辅助空间： O(N)