力扣刷题之101二叉树的对称判断

 此题写法可分为迭代和递归两种写法

1.递归法

首先明确判断二叉树对称是需要判断每个子树(而不是每个节点)的内侧，外侧是否相同，

引用自代码随想录的说明图

 

class Solution {
public:
    bool compare(TreeNode* left,TreeNode* right)
    {
        //跟NULL有关的节点进行判断
        if(left == NULL && right == NULL) return true;//均为空节点返回真
        if(left != NULL && right == NULL) return false;//有一节点不为空则说明不对称返回假
        if(left == NULL && right != NULL) return false;

        if(left->val != right -> val) return false;//两个均不为空但是值不相同不对称返回假

        //递归
        bool outside = compare(left->left,right->right);
        bool inside = compare(left->right,right->left);
        bool judge = outside && inside;

        return judge;
    }
    bool isSymmetric(TreeNode* root) 
    {
        return  compare(root->left,root->right);  
    }
};

2.迭代法

与递归法思路一致但是使用栈或队列来进行进一步判断

class Solution {
public:
    bool isSymmetric(TreeNode* root) 
    {
        queue<TreeNode*>queue1;
        if(root->left == NULL && root->right == NULL)
        {
            return true;
        }
        if(root->left == NULL || root ->right == NULL)
        {
            return false;
        }
        queue1.push(root->left);
        queue1.push(root->right);
        while(!queue1.empty())
        {
            TreeNode* left = queue1.front();
            queue1.pop();
            TreeNode* right = queue1.front();
            queue1.pop();

            if(!left && !right) continue;
            if(!left || !right || (left -> val != right ->val)) return false;
            queue1.push(left->left);
            queue1.push(right->right);
            queue1.push(left->right);
            queue1.push(right->left);
        }
        return true;
    }
};