将二叉树展开为链表 | 系列 2
原文：https://www.geeksforgeeks.org/flatten-a-binary-tree-into-linked-list-set-2/
给定一个二叉树，将其展开为一个链表。 展开后，每个节点的左侧应指向NULL，右侧应按级别顺序包含下一个节点。
示例：
Input:
1
/ \
2 5
/ \ \
3 4 6
Output:
1
\
2
\
3
\
4
\
5
\
6
Input:
1
/ \
3 4
/
2
\
5
Output:
1
\
3
\
4
\
2
\
5
方法：在先前的文章中已经讨论了使用递归的方法。 这种方法暗示了使用栈对二叉树进行预遍历。 在此遍历中，每次将右子项推入栈时，都会使右子项等于左子项，并使左子项等于NULL。 如果节点的右子节点变为NULL，则会弹出栈，而右子节点将从栈中弹出。 重复上述步骤，直到栈的大小为零或根为NULL。
下面是上述方法的实现：
C++
// C++ program to flatten the linked
// list using stack | set-2
#include <iostream>
#include <stack>
using namespace std;
struct Node {
int key;
Node *left, *right;
};
/* utility that allocates a new Node
with the given key */
Node* newNode(int key)
{
Node* node = new Node;
node->key = key;
node->left = node->right = NULL;
return (node);
}
// To find the inorder traversal
void inorder(struct Node* root)
{
// base condition
if (root == NULL)
return;
inorder(root->left);
cout << root->key << " ";
inorder(root->right);
}
// Function to convert binary tree into
// linked list by altering the right node
// and making left node point to NULL
Node* solution(Node* A)
{
// Declare a stack
stack<Node*> st;
Node* ans = A;
// Iterate till the stack is not empty
// and till root is Null
while (A != NULL || st.size() != 0) {
// Check for NULL
if (A->right != NULL) {
st.push(A->right);
}
// Make the Right Left and
// left NULL
A->right = A->left;
A->left = NULL;
// Check for NULL
if (A->right == NULL && st.size() != 0) {
A->right = st.top();
st.pop();
}
// Iterate
A = A->right;
}
return ans;
}
// Driver Code
int main()
{
/* 1
/ \
2 5
/ \ \
3 4 6 */
// Build the tree
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(5);
root->left->left = newNode(3);
root->left->right = newNode(4);
root->right->right = newNode(6);
// Call the function to
// flatten the tree
root = solution(root);
cout << "The Inorder traversal after "
"flattening binary tree ";
// call the function to print
// inorder after flatenning
inorder(root);
return 0;
return 0;
}
Java
// Java program to flatten the linked
// list using stack | set-2
import java.util.Stack;
class GFG
{
static class Node
{
int key;
Node left, right;
}
/* utility that allocates a new Node
with the given key */
static Node newNode(int key)
{
Node node = new Node();
node.key = key;
node.left = node.right = null;
return (node);
}
// To find the inorder traversal
static void inorder(Node root)
{
// base condition
if (root == null)
return;
inorder(root.left);
System.out.print(root.key + " ");
inorder(root.right);
}
// Function to convert binary tree into
// linked list by altering the right node
// and making left node point to null
static Node solution(Node A)
{
// Declare a stack
Stack<Node> st = new Stack<>();
Node ans = A;
// Iterate till the stack is not empty
// and till root is Null
while (A != null || st.size() != 0)
{
// Check for null
if (A.right != null)
{
st.push(A.right);
}
// Make the Right Left and
// left null
A.right = A.left;
A.left = null;
// Check for null
if (A.right == null && st.size() != 0)
{
A.right = st.peek();
st.pop();
}
// Iterate
A = A.right;
}
return ans;
}
// Driver Code
public static void main(String[] args)
{
/* 1
/ \
2 5
/ \ \
3 4 6 */
// Build the tree
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(5);
root.left.left = newNode(3);
root.left.right = newNode(4);
root.right.right = newNode(6);
// Call the function to
// flatten the tree
root = solution(root);
System.out.print("The Inorder traversal after "
+"flattening binary tree ");
// call the function to print
// inorder after flatenning
inorder(root);
}
}
// This code has been contributed by 29AjayKumar
C
// C# program to flatten the linked
// list using stack | set-2
using System;
using System.Collections.Generic;
class GFG
{
public class Node
{
public int key;
public Node left, right;
}
/* utility that allocates a new Node
with the given key */
static Node newNode(int key)
{
Node node = new Node();
node.key = key;
node.left = node.right = null;
return (node);
}
// To find the inorder traversal
static void inorder(Node root)
{
// base condition
if (root == null)
return;
inorder(root.left);
Console.Write(root.key + " ");
inorder(root.right);
}
// Function to convert binary tree into
// linked list by altering the right node
// and making left node point to null
static Node solution(Node A)
{
// Declare a stack
Stack<Node> st = new Stack<Node>();
Node ans = A;
// Iterate till the stack is not empty
// and till root is Null
while (A != null || st.Count != 0)
{
// Check for null
if (A.right != null)
{
st.Push(A.right);
}
// Make the Right Left and
// left null
A.right = A.left;
A.left = null;
// Check for null
if (A.right == null && st.Count != 0)
{
A.right = st.Peek();
st.Pop();
}
// Iterate
A = A.right;
}
return ans;
}
// Driver Code
public static void Main(String[] args)
{
/* 1
/ \
2 5
/ \ \
3 4 6 */
// Build the tree
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(5);
root.left.left = newNode(3);
root.left.right = newNode(4);
root.right.right = newNode(6);
// Call the function to
// flatten the tree
root = solution(root);
Console.Write("The Inorder traversal after "
+"flattening binary tree ");
// call the function to print
// inorder after flatenning
inorder(root);
}
}
// This code contributed by Rajput-Ji
输出：
The Inorder traversal after flattening binary tree 1 2 3 4 5 6
时间复杂度：O(n)
辅助空间：O(Log N)
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