给定一棵二叉树，检查它是否是它自己的镜像。
例子：
Input:
5
/ \
3 3
/ \ / \
8 9 9 8
Output: Symmetric
Input:
5
/ \
8 7
\ \
4 3
Output: Not Symmetric
方法：思想是使用 Morris Traversal 和 Reverse Morris Traversal 遍历树来遍历给定的二叉树，并在每一步检查当前节点的数据在两次遍历中是否相等。如果在任何步骤节点的数据不同。那么，给定的树不是对称二叉树。
下面是上述方法的实现：
C++
// C++ implementation to check
// if Tree is symmetric or not
#include
using namespace std;
// A Binary Tree Node
struct Node {
int data;
Node* left;
Node* right;
Node(int val)
{
data = val;
left = right = NULL;
}
};
// Function to check if the given
// binary tree is Symmetric or not
bool isSymmetric(struct Node* root)
{
Node *curr1 = root, *curr2 = root;
// Loop to traverse the tree in
// Morris Traversal and
// Reverse Morris Traversal
while (curr1 != NULL
&& curr2 != NULL) {
if (curr1->left == NULL
&& curr2->right == NULL) {
if (curr1->data != curr2->data)
return false;
curr1 = curr1->right;
curr2 = curr2->left;
}
else if (curr1->left != NULL
&& curr2->right != NULL) {
Node* pre1 = curr1->left;
Node* pre2 = curr2->right;
while (pre1->right != NULL
&& pre1->right != curr1
&& pre2->left != NULL
&& pre2->left != curr2) {
pre1 = pre1->right;
pre2 = pre2->left;
}
if (pre1->right == NULL
&& pre2->left == NULL) {
// Here, we are threading the Node
pre1->right = curr1;
pre2->left = curr2;
curr1 = curr1->left;
curr2 = curr2->right;
}
else if (pre1->right == curr1
&& pre2->left == curr2) {
// Unthreading the nodes
pre1->right = NULL;
pre2->left = NULL;
if (curr1->data != curr2->data)
return false;
curr1 = curr1->right;
curr2 = curr2->left;
}
else
return false;
}
else
return false;
}
if (curr1 != curr2)
return false;
return true;
}
// Driver Code
int main()
{
/*
5
/ \
3 3
/ \ / \
8 9 9 8 */
// Creation of Binary tree
Node* root = new Node(5);
root->left = new Node(3);
root->right = new Node(3);
root->left->left = new Node(8);
root->left->right = new Node(9);
root->right->left = new Node(9);
root->right->right = new Node(8);
if (isSymmetric(root))
cout << "Symmetric";
else
cout << "Not Symmetric";
return 0;
} Java
// Java implementation to check
// if Tree is symmetric or not
class GFG{
// A Binary Tree Node
static class Node
{
int data;
Node left;
Node right;
Node(int val)
{
data = val;
left = right = null;
}
};
// Function to check if the given
// binary tree is Symmetric or not
static boolean isSymmetric(Node root)
{
Node curr1 = root, curr2 = root;
// Loop to traverse the tree in
// Morris Traversal and
// Reverse Morris Traversal
while (curr1 != null &&
curr2 != null)
{
if (curr1.left == null &&
curr2.right == null)
{
if (curr1.data != curr2.data)
return false;
curr1 = curr1.right;
curr2 = curr2.left;
}
else if (curr1.left != null &&
curr2.right != null)
{
Node pre1 = curr1.left;
Node pre2 = curr2.right;
while (pre1.right != null &&
pre1.right != curr1 &&
pre2.left != null &&
pre2.left != curr2)
{
pre1 = pre1.right;
pre2 = pre2.left;
}
if (pre1.right == null &&
pre2.left == null)
{
// Here, we are threading the Node
pre1.right = curr1;
pre2.left = curr2;
curr1 = curr1.left;
curr2 = curr2.right;
}
else if (pre1.right == curr1 &&
pre2.left == curr2)
{
// Unthreading the nodes
pre1.right = null;
pre2.left = null;
if (curr1.data != curr2.data)
return false;
curr1 = curr1.right;
curr2 = curr2.left;
}
else
return false;
}
else
return false;
}
if (curr1 != curr2)
return false;
return true;
}
// Driver Code
public static void main(String[] args)
{
/*
5
/ \
3 3
/ \ / \
8 9 9 8 */
// Creation of Binary tree
Node root = new Node(5);
root.left = new Node(3);
root.right = new Node(3);
root.left.left = new Node(8);
root.left.right = new Node(9);
root.right.left = new Node(9);
root.right.right = new Node(8);
if (isSymmetric(root))
System.out.print("Symmetric");
else
System.out.print("Not Symmetric");
}
}
// This code is contributed by Rajput-JiPython3
# Python3 implementation to check
# if Tree is symmetric or not
# A Binary Tree Node
class Node:
def __init__(self, val):
self.data = val
self.left = self.right = None
# Function to check if the given
# binary tree is Symmetric or not
def isSymmetric(root):
curr1 = root
curr2 = root
# Loop to traverse the tree in
# Morris Traversal and
# Reverse Morris Traversal
while curr1 != None and curr2 != None:
if (curr1.left == None and
curr2.right == None):
if curr1.data != curr2.data:
return False
curr1 = curr1.right
curr2 = curr2.left
elif curr1 != None and curr2 != None:
pre1 = curr1.left
pre2 = curr2.right
while (pre1.right != None and
pre1.right != curr1 and
pre2.left != None and
pre2.left != curr2):
pre1 = pre1.right
pre2 = pre2.left
if pre1.right == None and pre2.left == None:
# Here, we are threading the Node
pre1.right = curr1
pre2.left = curr2
curr1 = curr1.left
curr2 = curr2.right
elif (pre1.right == curr1 and
pre2.left == curr2):
# Unthreading the nodes
pre1.right = None
pre2.left = None
if curr1.data != curr2.data:
return False
curr1 = curr1.right
curr2 = curr2.left
else:
return False
else:
return False
if curr1 != curr2:
return False
return True
# Driver code
def main():
#
# 5
# / \
# 3 3
# / \ / \
#8 9 9 8
# Creation of Binary tree
root = Node(5)
root.left = Node(3)
root.right = Node(3)
root.left.left = Node(8)
root.left.right = Node(9)
root.right.left = Node(9)
root.right.right = Node(8)
if isSymmetric(root):
print("Symmetric")
else:
print("Not Symmetric")
main()
# This code is contributed by Stuti PathakC#
// C# implementation to check
// if Tree is symmetric or not
using System;
class GFG{
// A Binary Tree Node
class Node
{
public int data;
public Node left;
public Node right;
public Node(int val)
{
data = val;
left = right = null;
}
};
// Function to check if the given
// binary tree is Symmetric or not
static bool isSymmetric(Node root)
{
Node curr1 = root, curr2 = root;
// Loop to traverse the tree in
// Morris Traversal and
// Reverse Morris Traversal
while (curr1 != null &&
curr2 != null)
{
if (curr1.left == null &&
curr2.right == null)
{
if (curr1.data != curr2.data)
return false;
curr1 = curr1.right;
curr2 = curr2.left;
}
else if (curr1.left != null &&
curr2.right != null)
{
Node pre1 = curr1.left;
Node pre2 = curr2.right;
while (pre1.right != null &&
pre1.right != curr1 &&
pre2.left != null &&
pre2.left != curr2)
{
pre1 = pre1.right;
pre2 = pre2.left;
}
if (pre1.right == null &&
pre2.left == null)
{
// Here, we are threading the Node
pre1.right = curr1;
pre2.left = curr2;
curr1 = curr1.left;
curr2 = curr2.right;
}
else if (pre1.right == curr1 &&
pre2.left == curr2)
{
// Unthreading the nodes
pre1.right = null;
pre2.left = null;
if (curr1.data != curr2.data)
return false;
curr1 = curr1.right;
curr2 = curr2.left;
}
else
return false;
}
else
return false;
}
if (curr1 != curr2)
return false;
return true;
}
// Driver Code
public static void Main(String[] args)
{
/*
5
/ \
3 3
/ \ / \
8 9 9 8 */
// Creation of Binary tree
Node root = new Node(5);
root.left = new Node(3);
root.right = new Node(3);
root.left.left = new Node(8);
root.left.right = new Node(9);
root.right.left = new Node(9);
root.right.right = new Node(8);
if (isSymmetric(root))
Console.Write("Symmetric");
else
Console.Write("Not Symmetric");
}
}
// This code is contributed by Rajput-Ji输出：
Symmetric
时间复杂度：由于树的每条边最多被遍历两次，就像莫里斯遍历的情况一样，在最坏的情况下，创建和删除相同数量的额外边(作为输入树)。因此，这种方法的时间复杂度是O(N) 。
辅助空间： O(1)
如果您想与行业专家一起参加直播课程，请参阅Geeks Classes Live