给定一个二叉树和一个由要删除的节点值组成的数组arr[] ，任务是在删除节点后打印森林的中序遍历。
例子：
Input: arr[] = {10, 5}
Output:
4 20
30 7
Input: arr[] = {5}
Output:
10
1 6 12
处理方法：按照以下步骤解决问题：
- 执行二叉树的后序遍历。
- 对于每个节点，检查它是否包含要删除的值。
- 如果发现为真，则将其子项存储为森林的根。通过中序遍历打印森林。
下面是上述方法的实现：
C++
10
/ \
20 30
/ \ \
4 5 7Java
1
/ \
5 6
/ \
10 12Python3
// C++ Program to implement
// the above approach
#include
using namespace std;
// Stores the nodes to be deleted
unordered_map mp;
// Structure of a Tree node
struct Node {
int key;
struct Node *left, *right;
};
// Function to create a new node
Node* newNode(int key)
{
Node* temp = new Node;
temp->key = key;
temp->left = temp->right = NULL;
return (temp);
}
// Function to check whether the node
// needs to be deleted or not
bool deleteNode(int nodeVal)
{
return mp.find(nodeVal) != mp.end();
}
// Function to perform tree pruning
// by performing post order traversal
Node* treePruning(Node* root, vector& result)
{
if (root == NULL)
return NULL;
root->left = treePruning(root->left, result);
root->right = treePruning(root->right, result);
// If the node needs to be deleted
if (deleteNode(root->key)) {
// Store the its subtree
if (root->left) {
result.push_back(root->left);
}
if (root->right) {
result.push_back(root->right);
}
return NULL;
}
return root;
}
// Perform Inorder Traversal
void printInorderTree(Node* root)
{
if (root == NULL)
return;
printInorderTree(root->left);
cout << root->key << " ";
printInorderTree(root->right);
}
// Function to print the forests
void printForests(Node* root, int arr[], int n)
{
for (int i = 0; i < n; i++) {
mp[arr[i]] = true;
}
// Stores the remaining nodes
vector result;
if (treePruning(root, result))
result.push_back(root);
// Print the inorder traversal of Forests
for (int i = 0; i < result.size(); i++) {
printInorderTree(result[i]);
cout << endl;
}
}
// Driver Code
int main()
{
Node* root = newNode(1);
root->left = newNode(12);
root->right = newNode(13);
root->right->left = newNode(14);
root->right->right = newNode(15);
root->right->left->left = newNode(21);
root->right->left->right = newNode(22);
root->right->right->left = newNode(23);
root->right->right->right = newNode(24);
int arr[] = { 14, 23, 1 };
int n = sizeof(arr) / sizeof(arr[0]);
printForests(root, arr, n);
} C#
// Java Program to implement
// the above approach
import java.util.*;
class GFG{
// Stores the nodes to be deleted
static HashMap mp = new HashMap<>();
// Structure of a Tree node
static class Node
{
int key;
Node left, right;
};
// Function to create a new node
static Node newNode(int key)
{
Node temp = new Node();
temp.key = key;
temp.left = temp.right = null;
return (temp);
}
// Function to check whether the node
// needs to be deleted or not
static boolean deleteNode(int nodeVal)
{
return mp.containsKey(nodeVal);
}
// Function to perform tree pruning
// by performing post order traversal
static Node treePruning(Node root,
Vector result)
{
if (root == null)
return null;
root.left = treePruning(root.left, result);
root.right = treePruning(root.right, result);
// If the node needs to be deleted
if (deleteNode(root.key))
{
// Store the its subtree
if (root.left != null)
{
result.add(root.left);
}
if (root.right != null)
{
result.add(root.right);
}
return null;
}
return root;
}
// Perform Inorder Traversal
static void printInorderTree(Node root)
{
if (root == null)
return;
printInorderTree(root.left);
System.out.print(root.key + " ");
printInorderTree(root.right);
}
// Function to print the forests
static void printForests(Node root,
int arr[], int n)
{
for (int i = 0; i < n; i++)
{
mp.put(arr[i], true);
}
// Stores the remaining nodes
Vector result = new Vector<>();
if (treePruning(root, result) != null)
result.add(root);
// Print the inorder traversal of Forests
for (int i = 0; i < result.size(); i++)
{
printInorderTree(result.get(i));
System.out.println();
}
}
// Driver Code
public static void main(String[] args)
{
Node root = newNode(1);
root.left = newNode(12);
root.right = newNode(13);
root.right.left = newNode(14);
root.right.right = newNode(15);
root.right.left.left = newNode(21);
root.right.left.right = newNode(22);
root.right.right.left = newNode(23);
root.right.right.right = newNode(24);
int arr[] = { 14, 23, 1 };
int n = arr.length;
printForests(root, arr, n);
}
}
// This code is contributed by Rajput-Ji 输出：
# Python3 Program to implement
# the above approach
# Stores the nodes to be deleted
mp = dict()
# Structure of a Tree node
class Node:
def __init__(self, key):
self.key = key
self.left = None
self.right = None
# Function to create a new node
def newNode(key):
temp = Node(key)
return temp
# Function to check whether the node
# needs to be deleted or not
def deleteNode(nodeVal):
if nodeVal in mp:
return True
else:
return False
# Function to perform tree pruning
# by performing post order traversal
def treePruning( root, result):
if (root == None):
return None;
root.left = treePruning(root.left, result);
root.right = treePruning(root.right, result);
# If the node needs to be deleted
if (deleteNode(root.key)):
# Store the its subtree
if (root.left):
result.append(root.left);
if (root.right):
result.append(root.right);
return None;
return root;
# Perform Inorder Traversal
def printInorderTree(root):
if (root == None):
return;
printInorderTree(root.left);
print(root.key, end=' ')
printInorderTree(root.right);
# Function to print the forests
def printForests(root, arr, n):
for i in range(n):
mp[arr[i]] = True;
# Stores the remaining nodes
result = []
if (treePruning(root, result)):
result.append(root);
# Print the inorder traversal of Forests
for i in range(len(result)):
printInorderTree(result[i]);
print()
# Driver Code
if __name__=='__main__':
root = newNode(1);
root.left = newNode(12);
root.right = newNode(13);
root.right.left = newNode(14);
root.right.right = newNode(15);
root.right.left.left = newNode(21);
root.right.left.right = newNode(22);
root.right.right.left = newNode(23);
root.right.right.right = newNode(24);
arr = [ 14, 23, 1 ]
n = len(arr)
printForests(root, arr, n);
# This code is contributed by rutvik_56时间复杂度： O(N)
辅助空间： O(1)
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