Problem statement:

问题陈述：

Given a binary Tree, check whether the tree is symmetric or not.

给定二叉树 ， 检查树是否对称 。

Input Example:

输入示例：

For example
1
/ \
2   2
/ \ / \
3  4 4  3
The above tree is symmetric
1
/ \
2   2
\   \
4    4
The above tree is not

Solution

解

对称树 (Symmetric tree)

A binary tree is said to be symmetric, if a vertical axis through the root cuts the tree in two mirror subtrees. That means the right child & left child of root are two mirror trees.

如果通过根的垂直轴在两个镜像子树中切割树，则二叉树被称为对称树。 这意味着根的右子和左子是两棵镜像树。

Algorithm:

算法：

Earlier, we have discussed how two check mirror trees? Kindly refer there for more detailed analysis.

前面我们讨论了如何检查两个镜像树 ？ 请参考那里进行更详细的分析。

In this case we are to check whether the tree is symmetric or not, which can be done via checking whether root's children are mirror trees or not?

在这种情况下，我们要检查树是否对称 ，这可以通过检查根的子节点是否为镜像树来完成？

FUNCTION isSymmetric(Tree Node* root) 
1.  Check for base cases
IF root==NULL //if root is NULL
Return true;
IF root->left==NULL &&root->right==NULL //if both child is NULL
Return true;
IF root->left==NULL || root->right==NULL //if only one child is NULL
Return false;
2.  check whether left & right subtrees are mirror of each other or not 
Return check(root->left,root->right);
END FUNCTION

FUNCTION Check(Tree Node* root1, Tree Node* root2)
a.  Check base cases
If root1==NULL && root2==NULL
Then it's mirror tree, return true;
If root1==NULL || root2==NULL
Then it's not a mirror tree, return false
Because one root is NULL whether another is not. 
(Both can't be NULL here, since already checked before) 
If root1->data != root2->data
Then it's not a mirror tree, return false.
Because roots are different & thus can't be mirror image of other
b.  Recursively check for sub-trees
Return (Check (root1->left, root2->right) &&Check(root1->right,root2->left));
END FUNCTION

Example with explanation

带说明的例子

For the example:
1
/ \
2   2
/ \ / \
3  4 4  3
At the main function:
It calls isSymmetric(1)
-----------------------------------------------------------
isSymmetric(1):
Base cases are not met
It calls check(1->left, 1->right)
Root1=2(1->left) and Root2=2(1->right)
It calls check (2, 2)
-----------------------------------------------------------
check (2, 2):
2->left =3 and 2->right=4 in case of tree1
2->left =4 and 2->right=3 in case of tree2
No base cases are satisfied thus it returns,
(check ( 3, 3) && check ( 4, 4))
Call to check ( 3, 3) and check ( 4, 4)
-----------------------------------------------------------
check (3, 3):(call at check (2, 2))
3->left =NULL and 3->right=NULL in case of tree1
3->left =NULL and 3->right=NULL in case of tree2
No base cases are satisfied thus it returns,
(check (NULL, NULL) &&check (NULL, NULL))
Call to check (NULL, NULL) and check (NULL, NULL))
-----------------------------------------------------------
check (4, 4): (call at check (2, 2))
4->left =NULL and 4->right=NULL in case of tree1
4->left =NULL and 4->right=NULL in case of tree2
No base cases are satisfied thus it returns,
(check (NULL, NULL) &&check (NULL, NULL))
Call to check (NULL, NULL) &&check (NULL, NULL)
-----------------------------------------------------------
check (NULL, NULL): (call at check (3, 3))
Base case matches and returns 1. 
-----------------------------------------------------------
check (NULL, NULL): (call at check (3, 3))
Base case matches and returns 1. 
-----------------------------------------------------------
check (NULL, NULL): (call at check (4, 4))
Base case matches and returns 1. 
-----------------------------------------------------------
check (NULL, NULL): (call at check (4, 4))
Base case matches and returns 1. 
-----------------------------------------------------------
Thus at isSymmetric(1),
check (2, 2) returns:
=   check ( 3, 3) &&check ( 4, 4)
=   ((check (NULL, NULL)&&check (NULL, NULL)) 
&&((check (NULL, NULL)&&check (NULL, NULL))
=   ((1 && 1)) && (1 && 1))
=   1
Thus at main it returns 1
So it's a symmetric tree

C++ implementation

C ++实现

#include<bits/stdc++.h>
using namespace std;
// TreeNode node type
class TreeNode{
public:             
int val;           //value
TreeNode *left;    //pointer to left child
TreeNode *right;   //pointer to right child
};
// creating new node
TreeNode* newnode(int data)  
{ 
TreeNode* node = (TreeNode*)malloc(sizeof(TreeNode)); 
node->val = data; 
node->left = NULL; 
node->right = NULL; 
return(node); 
}
// function to check mirror trees
bool check(TreeNode* root1,TreeNode* root2){
// base cases
//if both root is NULL, then it's mirror tree
if(!root1 && !root2)
return true;
//if one is NULL & other is not
//then it's not mirror tree
if(!root1 || !root2)
return false;
//if root data are different it's not mirror tree
if(root1->val!=root2->val)
return false;
//check for subtrees
return (check(root1->left,root2->right)&&check(root1->right,root2->left));
}
bool isSymmetric(TreeNode* root) {
//base cases
if(!root)
return true;
if(!root->left && !root->right)
return true;
if(!root->left || !root->right)
return false;
//check whether left & right subtrees 
//are mirror of each other or not 
return check(root->left,root->right);
}
int main(){
cout<<"tree is built as per example\n";
TreeNode *root=newnode(1); 
root->left= newnode(2); 
root->right= newnode(2); 
root->right->right=newnode(3);
root->right->left=newnode(4);
root->left->left=newnode(3); 
root->left->right=newnode(4);
if(isSymmetric(root))
cout<<"It's a symmetric tree\n";
else
cout<<"It's not a symmetric tree\n";
return 0;
}

Output

输出量

tree is built as per example
It's a symmetric tree 

翻译自: https://www.includehelp.com/icp/symmetric-tree.aspx