Leetcode: Symmetric Tree

题目：

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

Note:
Bonus points if you could solve it both recursively and iteratively.

思路分析：

题目要求分别用递归和迭代的方法进行。递归这里不说了，直接看代码。迭代算法可以使用C++标准库中的deque数据结构，允许从前面和后面操作元素。

递归算法：

class Solution{private:    bool checkSymmetric(TreeNode *left, TreeNode *right)    {    if (left == NULL && right == NULL)    {    return true;    }    else if (left && right && left->val == right->val)    {    return checkSymmetric(left->left, right->right) && checkSymmetric(left->right, right->left);    }    else    {    return false;    }    }public:    bool isSymmetric(TreeNode *root)    {    if (root == NULL)    {    return true;    }    else    {    return checkSymmetric(root->left, root->right);    }    }};

迭代算法（使用deque结构）：

class Solution{public:    bool isSymmetric(TreeNode *root)    {    if (!root)    {    return true;    }    //如果左右子树都为NULL    if (!root->left && !root->right)    {    return true;    }    //左右子树一个为NULL一个不为NULL    if ((root->left && !root->right) || (!root->left && root->right))    {    return false;    }    //左右子树都不为NULL    TreeNode *leftNode, *rightNode;    deque<TreeNode*> nodeDeque;nodeDeque.push_front(root->left);nodeDeque.push_back(root->right);while (!nodeDeque.empty()){leftNode = nodeDeque.front();rightNode = nodeDeque.back();nodeDeque.pop_front();nodeDeque.pop_back();if (leftNode->val != rightNode->val){return false;}if ((leftNode->left && !rightNode->right) || (!leftNode->left && rightNode->right)){return false;}if (leftNode->left){nodeDeque.push_front(leftNode->left);nodeDeque.push_back(rightNode->right);}if ((leftNode->right && !rightNode->left) || (!leftNode->right && rightNode->left)){return false;}if (leftNode->right){nodeDeque.push_front(leftNode->right);nodeDeque.push_back(rightNode->left);}}return true;}};

我看了下，使用递归算法在Leetcode上提交是8ms，使用迭代算法是10ms。

网上有的说采用中序遍历，判断遍历的结果对称就OK了。实际上这是不行的。

class Solution{private:    void inOrder(TreeNode *node, vector<int> &result)    {    if (node)    {        inOrder(node->left, result);    result.push_back(node->val);    inOrder(node->right, result);    }    }public:    bool isSymmetric(TreeNode *root)    {        if (root == NULL)        {            return true;        }    vector<int> result;    inOrder(root, result);    vector<int>::size_type size = result.size();    bool flag = true;    for (int i = 0, j = size - 1; i < j; i++, j--)    {    if (result[i] != result[j])    {    flag = false;    break;    }    }    return flag;    }};

这样的程序在比如{1， 2， 3， 3，#， 2}这样的测试用例就不能通过。