【剑指offer】对称的二叉树

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
*/
class Solution {
public:
    /*递归写法很简单，重写一个递归函数即可
    bool isSym(TreeNode* pRoot1, TreeNode* pRoot2){
        if(!pRoot1 && !pRoot2)
            return true;
        if(!pRoot1 || !pRoot2)
            return false;
        if(pRoot1->val != pRoot2->val)
            return false;
        return isSym(pRoot1->left, pRoot2->right) && isSym(pRoot1->right, pRoot2->left);
    }
    bool isSymmetrical(TreeNode* pRoot)
    {
        return isSym(pRoot, pRoot);
    }
    */
    //非递归写法DFS需要用到辅助栈,把每一对相同的点都成对加入到栈里
    //辅助空间利用栈和队列都可以，一个是DFS，一个是BFS
    bool isSymmetrical(TreeNode* pRoot)
    {
        stack<TreeNode* > Stack;
        if(!pRoot) return true;
        Stack.push(pRoot->left);
        Stack.push(pRoot->right);
        while(!Stack.empty()){
            TreeNode* right = Stack.top();
            Stack.pop();
            TreeNode* left = Stack.top();
            Stack.pop();
            if(left == nullptr && right == nullptr)
                continue;
            if(left == nullptr || right == nullptr)
                return false;
            if(left->val != right->val)
                return false;
            Stack.push(left->right);
            Stack.push(right->left);
            Stack.push(left->left);
            Stack.push(right->right);
        }
        return true;
    }
};