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

根开始，如果根节点的左右不对称，则false，否则，看根节点的左右子树是否对称。

代码：（AC）

/**
 * 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 left_right_Symmetric(TreeNode* pleft,TreeNode* pright)
{
     if(pleft==NULL&&pright==NULL) return true;
      if(pleft==NULL&&pright!=NULL)return false;
      if(pleft!=NULL&&pright==NULL)return false;
      if(pleft->val!=pright->val) return false;
      return left_right_Symmetric(pleft->left,pright->right)&&left_right_Symmetric(pleft->right,pright->left);
}
public:
    bool isSymmetric(TreeNode* root) {
         if(root == NULL) return true;
    if(root->left!=NULL&&root->right==NULL) return false;
    if(root->left==NULL&&root->right!=NULL) return false;
    if(root->left!=NULL&&root->right!=NULL&&root->left->val!=root->right->val)return false;
    else return left_right_Symmetric(root->left,root->right);
    }
};

错误代码：没有理解对称树的概念。（WA）

/**
 * 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==NULL) return true;
    if(root->left==NULL&&root->right!=NULL) return false;
    else if(root->left!=NULL&&root->right==NULL)return false;
         else if(root->left==NULL&&root->right==NULL) return true;
              else if(root->left!=NULL&&root->right!=NULL&&root->left->val!=root->right->val) return false;
                   else if(root->left!=NULL&&root->right!=NULL&&root->left->val==root->right->val)
                   return isSymmetric(root->left)&&isSymmetric(root->right);
    }
};