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 5The 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); } };