1.Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/
2 2
/ \ /
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/
2 2
\
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
2.題目要求：判斷二叉樹是否對稱，其實是判定兩棵樹是否鏡像欠窒。
3.方法：非遞歸酒繁，用棧來代替禽绪。因為是對稱比較帽驯，所以要兩個棧欠痴。這個思路其實可以稍微簡化一下尉桩，改用一個雙端隊列deque實現(xiàn)侈询。
4.代碼：
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;
}
};
