101. Symmetric Tree

二叉树对称性检查

最新推荐文章于 2021-06-11 11:16:20 发布

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#对称树 #递归 #非递归

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本文介绍了一种检查二叉树是否为自身镜像的方法，即判断其是否围绕中心对称。通过递归和非递归两种方式，详细解析了如何比较二叉树的左子树和右子树，以确定树的整体对称性。

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

But the following [1,2,2,null,3,null,3] is not:

Note:
Bonus points if you could solve it both recursively and iteratively.

方法1：递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root==NULL)
            return true;   
        return isSymmetric(root->left, root->right);
    }
private:
     bool isSymmetric(TreeNode* lNode,TreeNode* rNode)
     {
         if(lNode!=NULL && rNode!=NULL)
         {
             if(lNode->val != rNode->val)
                 return false;
             bool lres = isSymmetric(lNode->left, rNode->right);
             bool rres = isSymmetric(lNode->right, rNode->left);
             return lres && rres;                           
         }
        else if (lNode ==NULL && rNode == NULL)
            return true;
        else
            return false;  
     }       
};

方法2：非递归

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     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*> nodes;
       nodes.push(root->left);
       nodes.push(root->right);
       while(!nodes.empty())
       {   
           int len = nodes.size();
           while(len)
           {
               TreeNode* nodeL = nodes.front();
               nodes.pop();
               TreeNode* nodeR = nodes.front();              
               nodes.pop();
               len=len-2;
               if(nodeL && nodeR)
               {
                   if(nodeL->val != nodeR->val)
                    return false;
               } 
               else if(!nodeL && !nodeR)
               {
                   continue;
               }
               else 
               {
                   return false;
               }
               nodes.push(nodeL->left);
               nodes.push(nodeR->right);
               nodes.push(nodeL->right);
               nodes.push(nodeR->left);             
           }          
       } 
        return true;
    } 

};