[104]Symmetric Tree

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.
脑海中的我：按照中序左中右的顺序输出，然后另一个右中左的顺序输出，对比结果一样就皆大欢喜。然而，我好像想错了，比如上面第二棵树很明显就不行。
Solution1:（迭代）

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        queue<TreeNode*> st;
        st.push(root);          
        st.push(root); 
        {     
          while(!st.empty())
          {
            TreeNode* node1 = st.front();
            st.pop();
            TreeNode* node2 = st.front();
            st.pop();
            if(node1 == nullptr && node2 == nullptr) continue;
            else if((node1 == nullptr || node2==nullptr) || node1->val != node2->val)
            {
              return false;
            }
            st.push(node1->left);
            st.push(node2->right);               
            st.push(node1->right);
            st.push(node2->left);          
          }
        }
      return true;
    }
};

利用的树的层次遍历，往队列里面塞入树对称位置上的数，然后出队列时对比数字是否一致即可。

solution2:(递归)

/**
 * Definition for a binary tree node.
 * 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 == nullptr) return true;
        return isMirror(root->left,root->right);
    }
    bool isMirror(TreeNode* root1,TreeNode* root2)
    {
      if(root1 == nullptr && root2 == nullptr) return true;
      if(root1 == nullptr || root2 == nullptr) return false;
      return (root1->val == root2->val) && 
        isMirror(root1->left,root2->right)&&
        isMirror(root1->right,root2->left);
      
    }
};