[LeetCode]Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1 / \ 2 2 / \ / \ 3 4 4 3
But the following is not:
1 / \ 2 2 \ \ 3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/ \
2 3
/
4
\
5
The above binary tree is serialized as
"{1,2,3,#,#,4,#,#,5}".
思考：
递归：DFS，两个指针p、q。若p移到左子树，则q移到右子树，然后再比较。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool DFS(TreeNode *p,TreeNode *q)
{
if(!p&&!q) return true;
else if(p&&q)
{
if(p->val!=q->val) return false;
else return DFS(p->left,q->right)&&DFS(p->right,q->left);
}
else return false;
}
bool isSymmetric(TreeNode *root) {
TreeNode *p=root;
TreeNode *q=root;
return DFS(p,q);
}
};
另一种繁一点，先复制原树，然后翻转，再跟原树比较。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void copyandreverse(TreeNode *root,TreeNode *&newroot,TreeNode *p,bool flag)
{
if(root)
{
TreeNode *node=new TreeNode(root->val);
if(newroot==NULL)
{
newroot=p=node;
}
else
{
if(flag) p->right=node;
else p->left=node;
p=node;
}
copyandreverse(root->left,newroot,p,true);
copyandreverse(root->right,newroot,p,false);
}
}
bool DFS(TreeNode *root,TreeNode *newTree)
{
if(!root&&!newTree) return true;
else if(root&&newTree)
{
if(root->val!=newTree->val) return false;
else return DFS(root->left,newTree->left)&&DFS(root->right,newTree->right);
}
else return false;
}
bool isSymmetric(TreeNode *root) {
if(root==NULL) return true;
TreeNode *newTree=NULL;
TreeNode *p=newTree;
copyandreverse(root,newTree,p,true);
return DFS(root,newTree);
}
};