剑指offer28 对称的二叉树

给定一个二叉树，判断该二叉树是不是对称的。

1.递归方法

class Solution {
public:
    bool isSymmetrical(TreeNode* pRoot)
    {
        return isSymmetrical(pRoot,pRoot);
    }
    bool isSymmetrical(TreeNode* pRoot1,TreeNode* pRoot2){
        if(pRoot1==nullptr&&pRoot2==nullptr)
            return true;
        if(pRoot1==nullptr||pRoot2==nullptr)
            return false;
        if(pRoot1->val!=pRoot2->val)
            return false;
        return isSymmetrical(pRoot1->left,pRoot2->right)&&
            isSymmetrical(pRoot1->right,pRoot2->left);
    }
 
};

2.DFS:

利用栈，每次插入与取出都是左右同时插入。

class Solution {
public:
	bool isSymmetrical(TreeNode* pRoot)
	{
		if (pRoot == nullptr)
			return true;
		stack s;
		s.push(pRoot->left);
		s.push(pRoot->right);
		while (!s.empty()) {
			TreeNode* Right = s.top();
			s.pop();
			TreeNode* Left = s.top();
			s.pop();
			if (Right==nullptr && Left==nullptr)
				continue;
			if (Right==nullptr || Left==nullptr)
				return false;
			if (Right->val != Left->val)
				return false;
			s.push(Left->left);
			s.push(Right->right);
			s.push(Left->right);
			s.push(Right->left);
		}
		return true;
	}
};

3.BFS

和DFS差不多，不过这个是一层一层检查的，类似于层序遍历

class Solution {
public:
	bool isSymmetrical(TreeNode* pRoot)
	{
		if (pRoot == nullptr)
			return true;
		deque s;
		s.push_back(pRoot->left);
		s.push_back(pRoot->right);
		while (!s.empty()) {
			TreeNode* Right = s.front();
			s.pop_front();
			TreeNode* Left = s.front();
			s.pop_front();
			if (Right==nullptr && Left==nullptr)
				continue;
			if (Right==nullptr || Left==nullptr)
				return false;
			if (Right->val != Left->val)
				return false;
			s.push_back(Left->left);
			s.push_back(Right->right);
			s.push_back(Left->right);
			s.push_back(Right->left);
		}
		return true;
	}
};