给定两个二叉树 S 和 T，任务是检查 S 是否是树 T 的子树。
例如：
Input:
Tree T -
1
/ \
2 3
/ \ / \
4 5 6 7
Tree S -
2
/ \
4 5
Output: YES
Explanation:
The above tree is the subtree of the tree T,
Hence the output is YES
方法：这个想法是在 Pre-order Traversal 中遍历这棵树，并检查树的每个节点，该树的 Pre-order 遍历与树 S 的 Pre-order Traversal 相同，并注意 Null 值那是通过在遍历列表中包含空值，因为如果不注意空值，那么两个不同的树可以具有相同的前序遍历。例如在下面的树的情况下——
1 1
/ \ / \
2 N N 2
/ \ / \
N N N N
Pre-order Traversal of both the trees will be -
{1, 2} - In case of without taking care of null values
{1, 2, N, N, N} and {1, N, 2, N, N}
In case of taking care of null values.
算法：
- 声明一个堆栈，以跟踪节点的左右子节点。
- 压入树 T 的根节点。
- 在栈不为空的情况下运行一个循环，然后检查栈顶节点的前序遍历是否相同，则返回true。
- 如果前序遍历与树不匹配，则从堆栈中弹出顶部节点并推送其弹出节点的左右子节点。
下面是上述方法的实现：
C++
// C++ implementation to check
// if a tree is a subtree of
// another binary tree
#include
using namespace std;
// Structure of the
// binary tree node
struct Node {
int data;
struct Node* left;
struct Node* right;
};
// Function to create
// new node
Node* newNode(int x)
{
Node* temp = (Node*)malloc(
sizeof(Node));
temp->data = x;
temp->left = NULL;
temp->right = NULL;
return temp;
}
// Function to check if two trees
// have same pre-order traversal
bool areTreeIdentical(Node* t1, Node* t2)
{
stack s1;
stack s2;
Node* temp1;
Node* temp2;
s1.push(t1);
s2.push(t2);
// Loop to iterate over the stacks
while (!s1.empty() && !s2.empty()) {
temp1 = s1.top();
temp2 = s2.top();
s1.pop();
s2.pop();
// Both are None
// hence they are equal
if (temp1 == NULL &&
temp2 == NULL)
continue;
// nodes are uneuqal
if ((temp1 == NULL && temp2 != NULL) ||
(temp1 != NULL && temp2 == NULL))
return false;
// nodes have unqual data
if (temp1->data != temp2->data)
return false;
s1.push(temp1->right);
s2.push(temp2->right);
s1.push(temp1->left);
s2.push(temp2->left);
}
// if both tree are identical
// both stacks must be empty.
if (s1.empty() && s2.empty())
return true;
else
return false;
}
// Function to check if the Tree s
// is the subtree of the Tree T
bool isSubTree(Node* s, Node* t)
{
// first we find the root of s in t
// by traversing in pre order fashion
stack stk;
Node* temp;
stk.push(t);
while (!stk.empty()) {
temp = stk.top();
stk.pop();
// if current node data is equal
// to root of s then
if (temp->data == s->data) {
if (areTreeIdentical(s, temp))
return true;
}
if (temp->right)
stk.push(temp->right);
if (temp->left)
stk.push(temp->left);
}
return false;
}
// Driver Code
int main()
{
/*
1
/ \
2 3
/ \ / \
4 5 6 7
*/
Node* root = newNode(1);
root->left = newNode(2);
root->right = newNode(3);
root->left->left = newNode(4);
root->left->right = newNode(5);
root->right->left = newNode(6);
root->right->right = newNode(7);
/*
2
/ \
4 5
*/
Node* root2 = newNode(2);
root2->left = newNode(4);
root2->right = newNode(5);
if (isSubTree(root2, root))
cout << "Yes";
else
cout << "No";
return 0;
} Java
// Java implementation to check
// if a tree is a subtree of
// another binary tree
import java.util.*;
class GFG{
// Structure of the
// binary tree node
static class Node {
int data;
Node left;
Node right;
};
// Function to create
// new node
static Node newNode(int x)
{
Node temp = new Node();
temp.data = x;
temp.left = null;
temp.right = null;
return temp;
}
// Function to check if two trees
// have same pre-order traversal
static boolean areTreeIdentical(Node t1, Node t2)
{
Stack s1 = new Stack();
Stack s2 = new Stack();
Node temp1;
Node temp2;
s1.add(t1);
s2.add(t2);
// Loop to iterate over the stacks
while (!s1.isEmpty() && !s2.isEmpty()) {
temp1 = s1.peek();
temp2 = s2.peek();
s1.pop();
s2.pop();
// Both are None
// hence they are equal
if (temp1 == null &&
temp2 == null)
continue;
// nodes are uneuqal
if ((temp1 == null && temp2 != null) ||
(temp1 != null && temp2 == null))
return false;
// nodes have unqual data
if (temp1.data != temp2.data)
return false;
s1.add(temp1.right);
s2.add(temp2.right);
s1.add(temp1.left);
s2.add(temp2.left);
}
// if both tree are identical
// both stacks must be empty.
if (s1.isEmpty() && s2.isEmpty())
return true;
else
return false;
}
// Function to check if the Tree s
// is the subtree of the Tree T
static boolean isSubTree(Node s, Node t)
{
// first we find the root of s in t
// by traversing in pre order fashion
Stack stk = new Stack();
Node temp;
stk.add(t);
while (!stk.isEmpty()) {
temp = stk.peek();
stk.pop();
// if current node data is equal
// to root of s then
if (temp.data == s.data) {
if (areTreeIdentical(s, temp))
return true;
}
if (temp.right != null)
stk.add(temp.right);
if (temp.left != null)
stk.add(temp.left);
}
return false;
}
// Driver Code
public static void main(String[] args)
{
/*
1
/ \
2 3
/ \ / \
4 5 6 7
*/
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(6);
root.right.right = newNode(7);
/*
2
/ \
4 5
*/
Node root2 = newNode(2);
root2.left = newNode(4);
root2.right = newNode(5);
if (isSubTree(root2, root))
System.out.print("Yes");
else
System.out.print("No");
}
}
// This code is contributed by Rajput-Ji Python3
# Python3 implementation to check
# if a tree is a subtree of
# another binary tree
# Structure of the
# binary tree node
class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Function to check if two trees
# have same pre-order traversal
def areTreeIdentical(t1 : Node,
t2 : Node) -> bool:
s1 = []
s2 = []
s1.append(t1)
s2.append(t2)
# Loop to iterate
# over the stacks
while s1 and s2:
temp1 = s1.pop()
temp2 = s2.pop()
# Both are None
# hence they are equal
if (temp1 is None and
temp2 is None):
continue
# Nodes are uneuqal
if ((temp1 is None and
temp2 is not None) or
(temp1 is not None and
temp2 is None)):
return False
# Nodes have unqual data
if (temp1.data != temp2.data):
return False
s1.append(temp1.right)
s2.append(temp2.right)
s1.append(temp1.left)
s2.append(temp2.left)
# If both tree are identical
# both stacks must be empty.
if s1 and s2:
return False
return True
# Function to check if the Tree s
# is the subtree of the Tree T
def isSubTree(s : Node,
t : Node) -> Node:
# First we find the
# root of s in t
# by traversing in
# pre order fashion
stk = []
stk.append(t)
while stk:
temp = stk.pop()
# If current node data is equal
# to root of s then
if (temp.data == s.data):
if (areTreeIdentical(s, temp)):
return True
if (temp.right):
stk.append(temp.right)
if (temp.left):
stk.append(temp.left)
return False
# Driver Code
if __name__ == "__main__":
'''
1
/ \
2 3
/ \ / \
4 5 6 7
'''
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
root.right.left = Node(6)
root.right.right = Node(7)
'''
2
/ \
4 5
'''
root2 = Node(2)
root2.left = Node(4)
root2.right = Node(5)
if (isSubTree(root2, root)):
print("Yes")
else:
print("No")
# This code is contributed by sanjeev2552C#
// C# implementation to check
// if a tree is a subtree of
// another binary tree
using System;
using System.Collections.Generic;
class GFG{
// Structure of the
// binary tree node
class Node {
public int data;
public Node left;
public Node right;
};
// Function to create
// new node
static Node newNode(int x)
{
Node temp = new Node();
temp.data = x;
temp.left = null;
temp.right = null;
return temp;
}
// Function to check if two trees
// have same pre-order traversal
static bool areTreeIdentical(Node t1, Node t2)
{
Stack s1 = new Stack();
Stack s2 = new Stack();
Node temp1;
Node temp2;
s1.Push(t1);
s2.Push(t2);
// Loop to iterate over the stacks
while (s1.Count != 0 && s2.Count != 0) {
temp1 = s1.Peek();
temp2 = s2.Peek();
s1.Pop();
s2.Pop();
// Both are None
// hence they are equal
if (temp1 == null &&
temp2 == null)
continue;
// nodes are uneuqal
if ((temp1 == null && temp2 != null) ||
(temp1 != null && temp2 == null))
return false;
// nodes have unqual data
if (temp1.data != temp2.data)
return false;
s1.Push(temp1.right);
s2.Push(temp2.right);
s1.Push(temp1.left);
s2.Push(temp2.left);
}
// if both tree are identical
// both stacks must be empty.
if (s1.Count == 0 && s2.Count == 0)
return true;
else
return false;
}
// Function to check if the Tree s
// is the subtree of the Tree T
static bool isSubTree(Node s, Node t)
{
// first we find the root of s in t
// by traversing in pre order fashion
Stack stk = new Stack();
Node temp;
stk.Push(t);
while (stk.Count != 0) {
temp = stk.Peek();
stk.Pop();
// if current node data is equal
// to root of s then
if (temp.data == s.data) {
if (areTreeIdentical(s, temp))
return true;
}
if (temp.right != null)
stk.Push(temp.right);
if (temp.left != null)
stk.Push(temp.left);
}
return false;
}
// Driver Code
public static void Main(String[] args)
{
/*
1
/ \
2 3
/ \ / \
4 5 6 7
*/
Node root = newNode(1);
root.left = newNode(2);
root.right = newNode(3);
root.left.left = newNode(4);
root.left.right = newNode(5);
root.right.left = newNode(6);
root.right.right = newNode(7);
/*
2
/ \
4 5
*/
Node root2 = newNode(2);
root2.left = newNode(4);
root2.right = newNode(5);
if (isSubTree(root2, root))
Console.Write("Yes");
else
Console.Write("No");
}
}
// This code is contributed by Princi Singh 输出：
Yes
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