检查一棵二叉树是否是另一棵二叉树的子树 |设置 1
给定两棵二叉树，检查第一棵树是否是第二棵树的子树。树 T 的子树是由 T 中的一个节点及其在 T 中的所有后代组成的树 S。根节点对应的子树就是整棵树；对应于任何其他节点的子树称为适当子树。
例如，在以下情况下，树 S 是树 T 的子树。
Tree 2
10
/ \
4 6
\
30Tree 1
26
/ \
10 3
/ \ \
4 6 3
\
30解决方案：以前序方式遍历树 T。对于遍历中的每个访问节点，查看以该节点为根的子树是否与 S 相同。
以下是对此的实现。
C++
// C++ program to check if binary tree
// is subtree of another binary tree
#include
using namespace std;
/* A binary tree node has data,
left child and right child */
class node
{
public:
int data;
node* left;
node* right;
};
/* A utility function to check
whether trees with roots as root1 and
root2 are identical or not */
bool areIdentical(node * root1, node *root2)
{
/* base cases */
if (root1 == NULL && root2 == NULL)
return true;
if (root1 == NULL || root2 == NULL)
return false;
/* Check if the data of both roots is
same and data of left and right
subtrees are also same */
return (root1->data == root2->data &&
areIdentical(root1->left, root2->left) &&
areIdentical(root1->right, root2->right) );
}
/* This function returns true if S
is a subtree of T, otherwise false */
bool isSubtree(node *T, node *S)
{
/* base cases */
if (S == NULL)
return true;
if (T == NULL)
return false;
/* Check the tree with root as current node */
if (areIdentical(T, S))
return true;
/* If the tree with root as current
node doesn't match then try left
and right subtrees one by one */
return isSubtree(T->left, S) ||
isSubtree(T->right, S);
}
/* Helper function that allocates
a new node with the given data
and NULL left and right pointers. */
node* newNode(int data)
{
node* Node = new node();
Node->data = data;
Node->left = NULL;
Node->right = NULL;
return(Node);
}
/* Driver code*/
int main()
{
// TREE 1
/* Construct the following tree
26
/ \
10 3
/ \ \
4 6 3
\
30
*/
node *T = newNode(26);
T->right = newNode(3);
T->right->right = newNode(3);
T->left = newNode(10);
T->left->left = newNode(4);
T->left->left->right = newNode(30);
T->left->right = newNode(6);
// TREE 2
/* Construct the following tree
10
/ \
4 6
\
30
*/
node *S = newNode(10);
S->right = newNode(6);
S->left = newNode(4);
S->left->right = newNode(30);
if (isSubtree(T, S))
cout << "Tree 2 is subtree of Tree 1";
else
cout << "Tree 2 is not a subtree of Tree 1";
return 0;
}
// This code is contributed by rathbhupendra C
#include
#include
/* A binary tree node has data, left child and right child */
struct node
{
int data;
struct node* left;
struct node* right;
};
/* A utility function to check whether trees with roots as root1 and
root2 are identical or not */
bool areIdentical(struct node * root1, struct node *root2)
{
/* base cases */
if (root1 == NULL && root2 == NULL)
return true;
if (root1 == NULL || root2 == NULL)
return false;
/* Check if the data of both roots is same and data of left and right
subtrees are also same */
return (root1->data == root2->data &&
areIdentical(root1->left, root2->left) &&
areIdentical(root1->right, root2->right) );
}
/* This function returns true if S is a subtree of T, otherwise false */
bool isSubtree(struct node *T, struct node *S)
{
/* base cases */
if (S == NULL)
return true;
if (T == NULL)
return false;
/* Check the tree with root as current node */
if (areIdentical(T, S))
return true;
/* If the tree with root as current node doesn't match then
try left and right subtrees one by one */
return isSubtree(T->left, S) ||
isSubtree(T->right, S);
}
/* Helper function that allocates a new node with the given data
and NULL left and right pointers. */
struct node* newNode(int data)
{
struct node* node =
(struct node*)malloc(sizeof(struct node));
node->data = data;
node->left = NULL;
node->right = NULL;
return(node);
}
/* Driver program to test above function */
int main()
{
// TREE 1
/* Construct the following tree
26
/ \
10 3
/ \ \
4 6 3
\
30
*/
struct node *T = newNode(26);
T->right = newNode(3);
T->right->right = newNode(3);
T->left = newNode(10);
T->left->left = newNode(4);
T->left->left->right = newNode(30);
T->left->right = newNode(6);
// TREE 2
/* Construct the following tree
10
/ \
4 6
\
30
*/
struct node *S = newNode(10);
S->right = newNode(6);
S->left = newNode(4);
S->left->right = newNode(30);
if (isSubtree(T, S))
printf("Tree 2 is subtree of Tree 1");
else
printf("Tree 2 is not a subtree of Tree 1");
getchar();
return 0;
} Java
// Java program to check if binary tree is subtree of another binary tree
// A binary tree node
class Node
{
int data;
Node left, right, nextRight;
Node(int item)
{
data = item;
left = right = nextRight = null;
}
}
class BinaryTree
{
Node root1,root2;
/* A utility function to check whether trees with roots as root1 and
root2 are identical or not */
boolean areIdentical(Node root1, Node root2)
{
/* base cases */
if (root1 == null && root2 == null)
return true;
if (root1 == null || root2 == null)
return false;
/* Check if the data of both roots is same and data of left and right
subtrees are also same */
return (root1.data == root2.data
&& areIdentical(root1.left, root2.left)
&& areIdentical(root1.right, root2.right));
}
/* This function returns true if S is a subtree of T, otherwise false */
boolean isSubtree(Node T, Node S)
{
/* base cases */
if (S == null)
return true;
if (T == null)
return false;
/* Check the tree with root as current node */
if (areIdentical(T, S))
return true;
/* If the tree with root as current node doesn't match then
try left and right subtrees one by one */
return isSubtree(T.left, S)
|| isSubtree(T.right, S);
}
public static void main(String args[])
{
BinaryTree tree = new BinaryTree();
// TREE 1
/* Construct the following tree
26
/ \
10 3
/ \ \
4 6 3
\
30 */
tree.root1 = new Node(26);
tree.root1.right = new Node(3);
tree.root1.right.right = new Node(3);
tree.root1.left = new Node(10);
tree.root1.left.left = new Node(4);
tree.root1.left.left.right = new Node(30);
tree.root1.left.right = new Node(6);
// TREE 2
/* Construct the following tree
10
/ \
4 6
\
30 */
tree.root2 = new Node(10);
tree.root2.right = new Node(6);
tree.root2.left = new Node(4);
tree.root2.left.right = new Node(30);
if (tree.isSubtree(tree.root1, tree.root2))
System.out.println("Tree 2 is subtree of Tree 1 ");
else
System.out.println("Tree 2 is not a subtree of Tree 1");
}
}
// This code has been contributed by Mayank JaiswalPython3
# Python program to check binary tree is a subtree of
# another tree
# A binary tree node
class Node:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# A utility function to check whether trees with roots
# as root 1 and root2 are indetical or not
def areIdentical(root1, root2):
# Base Case
if root1 is None and root2 is None:
return True
if root1 is None or root2 is None:
return False
# Check fi the data of both roots is same and data of
# left and right subtrees are also same
return (root1.data == root2.data and
areIdentical(root1.left , root2.left)and
areIdentical(root1.right, root2.right)
)
# This function returns True if S is a subtree of T,
# otherwise False
def isSubtree(T, S):
# Base Case
if S is None:
return True
if T is None:
return False
# Check the tree with root as current node
if (areIdentical(T, S)):
return True
# IF the tree with root as current node doesn't match
# then try left and right subtreee one by one
return isSubtree(T.left, S) or isSubtree(T.right, S)
# Driver program to test above function
""" TREE 1
Construct the following tree
26
/ \
10 3
/ \ \
4 6 3
\
30
"""
T = Node(26)
T.right = Node(3)
T.right.right = Node(3)
T.left = Node(10)
T.left.left = Node(4)
T.left.left.right = Node(30)
T.left.right = Node(6)
""" TREE 2
Construct the following tree
10
/ \
4 6
\
30
"""
S = Node(10)
S.right = Node(6)
S.left = Node(4)
S.left.right = Node(30)
if isSubtree(T, S):
print ("Tree 2 is subtree of Tree 1")
else :
print ("Tree 2 is not a subtree of Tree 1")
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)C#
// C# program to check if binary tree
// is subtree of another binary tree
using System;
// A binary tree node
class Node
{
public int data;
public Node left, right, nextRight;
public Node(int item)
{
data = item;
left = right = nextRight = null;
}
}
public class BinaryTree
{
Node root1,root2;
/* A utility function to check whether
trees with roots as root1 and
root2 are identical or not */
bool areIdentical(Node root1, Node root2)
{
/* base cases */
if (root1 == null && root2 == null)
return true;
if (root1 == null || root2 == null)
return false;
/* Check if the data of both roots is
same and data of left and right
subtrees are also same */
return (root1.data == root2.data
&& areIdentical(root1.left, root2.left)
&& areIdentical(root1.right, root2.right));
}
/* This function returns true if S is
a subtree of T, otherwise false */
bool isSubtree(Node T, Node S)
{
/* base cases */
if (S == null)
return true;
if (T == null)
return false;
/* Check the tree with root as current node */
if (areIdentical(T, S))
return true;
/* If the tree with root as current
node doesn't match then try left
and right subtrees one by one */
return isSubtree(T.left, S)
|| isSubtree(T.right, S);
}
// Driver code
public static void Main()
{
BinaryTree tree = new BinaryTree();
// TREE 1
/* Construct the following tree
26
/ \
10 3
/ \ \
4 6 3
\
30 */
tree.root1 = new Node(26);
tree.root1.right = new Node(3);
tree.root1.right.right = new Node(3);
tree.root1.left = new Node(10);
tree.root1.left.left = new Node(4);
tree.root1.left.left.right = new Node(30);
tree.root1.left.right = new Node(6);
// TREE 2
/* Construct the following tree
10
/ \
4 6
\
30 */
tree.root2 = new Node(10);
tree.root2.right = new Node(6);
tree.root2.left = new Node(4);
tree.root2.left.right = new Node(30);
if (tree.isSubtree(tree.root1, tree.root2))
Console.WriteLine("Tree 2 is subtree of Tree 1 ");
else
Console.WriteLine("Tree 2 is not a subtree of Tree 1");
}
}
/* This code is contributed by Rajput-Ji*/Javascript
输出：
Tree 2 is subtree of Tree 1 时间复杂度：上述解决方案的时间最坏情况复杂度为 O(mn)，其中 m 和 n 是给定两棵树中的节点数。
辅助空间： O(n)
我们可以在 O(n) 时间内解决上述问题。请参考 检查一棵二叉树是否是另一棵二叉树的子树 |为 O(n) 解决方案设置 2。