LeetCode之Symmetric Tree

</pre>Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).<p></p><p style="box-sizing: border-box; margin-top: 0px; margin-bottom: 10px; color: rgb(51, 51, 51); font-family: 'Helvetica Neue', Helvetica, Arial, sans-serif; font-size: 14px; line-height: 30px;">For example, this binary tree is symmetric:</p><pre style="box-sizing: border-box; overflow: auto; font-family: Menlo, Monaco, Consolas, 'Courier New', monospace; font-size: 13px; padding: 9.5px; margin-top: 0px; margin-bottom: 10px; line-height: 1.42857143; color: rgb(51, 51, 51); word-break: break-all; word-wrap: break-word; background-color: rgb(245, 245, 245); border: 1px solid rgb(204, 204, 204); border-top-left-radius: 4px; border-top-right-radius: 4px; border-bottom-right-radius: 4px; border-bottom-left-radius: 4px;">    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.

/**
 * 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);
    }
};


posted @ 2015-05-11 20:17  sunp823  阅读(116)  评论(0编辑  收藏  举报