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  3But the following is not:    1   / \  2   2   \   \   3    3

解答：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class NodePair {public:   NodePair(TreeNode* a, TreeNode* b) : pa(a), pb(b) {}   bool isMirror() {        bool a = (pa->left && pb->right && pa->left->val == pb->right->val) || (!pa->left && !pb->right);        bool b = (pb->left && pa->right && pb->left->val == pa->right->val) || (!pb->left && !pa->right);        return a && b;   }   TreeNode* pa;   TreeNode* pb;};class Solution {public:    bool isSymmetric(TreeNode* root) {        if (!root) {            return true;        }         NodePair rootPair(root, root);        if (!rootPair.isMirror()) {            return false;        }        if (!root->left) {            return true;        }        list<NodePair> myList;        NodePair pair(root->left, root->right);        myList.push_back(pair);        for (list<NodePair>::iterator it = myList.begin(); it != myList.end(); ++it) {            if(!it->isMirror()) {                return false;            }            if(it->pa->left) {                NodePair pair(it->pa->left, it->pb->right);                myList.push_back(pair);            }            if(it->pa->right) {                NodePair pair(it->pa->right, it->pb->left);                myList.push_back(pair);            }        }        return true;    }};