HappyLeetcode43: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

这道题目可以用迭代的方式做，也可以用递归的方式做。遇见和树相关的题目，第一反应是递归，所以我的解法是递归。

代码奉上：

class Solution {

 public:

     bool IsSymmetric(TreeNode *node1, TreeNode *node2)

     {

         if (node1 == NULL&&node2 == NULL)

             return true;

         if (node1 == NULL||node2 == NULL)

             return false;

         bool result1, result2;

         if (node1->val != node2->val)

             return false;

         else

         {

             result1 = IsSymmetric(node1->left, node2->right);

             result2 = IsSymmetric(node1->right, node2->left);

         }

         return result1&&result2;

     }

     bool isSymmetric(TreeNode *root) {

         if (root == NULL || root->left == NULL&&root->right == NULL)

             return true;

         bool result = IsSymmetric(root->left, root->right);

         return result;

     }

 };

看了相关人的解法，有的人直接用迭代的方式解出来了，感觉很受启发

/**

 * 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) {

        // IMPORTANT: Please reset any member data you declared, as

        // the same Solution instance will be reused for each test case.

        if(root == NULL)return true;

        queue<TreeNode*> qleft, qright;

        if(root->left)qleft.push(root->left);

        if(root->right)qright.push(root->right);

        while(qleft.empty() == false && qright.empty() == false)

        {

            TreeNode *ql = qleft.front();

            TreeNode *qr = qright.front();

            qleft.pop();  qright.pop();

            if(ql->val == qr->val)

            {

                if(ql->left && qr->right)

                {

                    qleft.push(ql->left);

                    qright.push(qr->right);

                }

                else if(ql->left || qr->right)

                    return false;

                if(qr->left && ql->right)

                {

                    qleft.push(qr->left);

                    qright.push(ql->right);

                }

                else if(qr->left || ql->right)

                    return false;

            }

            else return false;

        }

        if(qleft.empty() && qright.empty())

            return true;

        else return false;

    }

};

转自http://www.cnblogs.com/TenosDoIt/p/3440729.html