二叉树系列 - [LeetCode] Symmetric Tree 判断二叉树是否对称，递归和非递归实现

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.

二叉树是否对称的本质，其实是判定两棵树是否镜像。

递归是很常见的实现方式，最简便。

/**
* 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 isSymmetric(TreeNode *root) {
if(!root) return true;
return compRoot(root -> left, root -> right);
}
private:
bool compRoot(TreeNode* lroot, TreeNode* rroot){
if(!lroot) return (NULL == rroot);
if(NULL == rroot) return false;
if(lroot -> val != rroot -> val) return false;
return (compRoot(lroot -> left, rroot -> right) && compRoot(lroot -> right, rroot -> left));
}
};

非递归，我的方法其实还是很常规，用栈来代替。因为是对称比较，所以要两个栈。

这个思路其实可以稍微简化一下，改用一个双端队列deque实现。比起用两个栈来，显得稍微“洋气”一点 ==。

/**
* 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 isSymmetric(TreeNode *root) {
if(!root) return true;
if(!root -> left && !root -> right) return true;
if( (!root -> left && root -> right) || (root -> left && !root -> right) ) return false;
deque<TreeNode*> dq;
dq.push_front(root -> left);
dq.push_back(root -> right);
while(!dq.empty()){
TreeNode* lroot = dq.front();
TreeNode* rroot = dq.back();
dq.pop_front();
dq.pop_back();
if(lroot -> val != rroot -> val) return false;
if( (!lroot -> right && rroot -> left) || (lroot -> right && !rroot -> left) ) return false;
if(lroot -> right){
dq.push_front(lroot -> right);
dq.push_back(rroot -> left);
}
if( (!lroot -> left && rroot -> right) || (lroot -> left && !rroot -> right) ) return false;
if(lroot -> left){
dq.push_front(lroot -> left);
dq.push_back(rroot -> right);
}
}
return true;
}
};