题目描述:请实现一个函数,用来判断一颗二叉树是不是对称的。注意,如果一个二叉树同此二叉树的镜像是同样的,定义其为对称的。

    	    A
    	   /  \
    	  B    B
    	 / \  / \
    	C  D D   C

C++代码:

//普通递归方法
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};
class Solution {
public:
	bool isSymmetrical(TreeNode* pRoot)
	{
		if (pRoot == NULL)return true;
		return isSymmetrical(pRoot, pRoot);
	}
	bool isSymmetrical(TreeNode* pRoot1, TreeNode* pRoot2)
	{
			if (pRoot1 == NULL && pRoot2 == NULL)
				return true;
			if ((pRoot1 == NULL || pRoot2 == NULL))return false;
			if (pRoot1->val != pRoot2->val)return false;
			return isSymmetrical(pRoot1->left, pRoot2->right) && isSymmetrical(pRoot1->right, pRoot2->left);
	}
};

 

//使用队列
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root==NULL) return true;
        queue<TreeNode*> q1,q2;
        TreeNode *left,*right;
        q1.push(root->left);
        q2.push(root->right);
        while(!q1.empty() and !q2.empty())
        {
            left = q1.front();
            q1.pop();
            right = q2.front();
            q2.pop();
            //两边都是空
            if(NULL==left && NULL==right)
                continue;
            //只有一边是空
            if(NULL==left||NULL==right)
                return false;
             if (left->val != right->val)
                return false;
            q1.push(left->left);
            q1.push(left->right);
            q2.push(right->right);
            q2.push(right->left);
        }
         
        return true;
         
    }
};

 

 

//使用栈
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};

class Solution {
public:

    bool isSymmetrical(TreeNode* pRoot)
    {
        stack<TreeNode*> s1,s2;
        TreeNode *p1,*p2;
        p1=p2=pRoot;
        while((!s1.empty()&&!s2.empty())||(p1!=NULL&&p2!=NULL)){
            while(p1!=NULL&&p2!=NULL){
                s1.push(p1);
                s2.push(p2);
                p1=p1->left;
                p2=p2->right;
            }
            p1=s1.top();
            s1.pop();
            p2=s2.top();
            s2.pop();
            if(p1->val!=p2->val)
                return false;
            p1=p1->right;
            p2=p2->left;
        }
        if(!s1.empty()||!s2.empty())
            return false;
        if(p1!=NULL||p2!=NULL)
            return false;
        return true;
    }
};