LeetCode *** 101. Symmetric Tree

最新推荐文章于 2023-09-05 08:07:52 发布

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本文介绍了一种检查二叉树是否为镜面对称的方法，通过创建原树的镜像副本并比较两棵树的结构来判断原树是否对称。提供了一段使用C++实现的代码示例。

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题目：

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

For example, this binary tree is symmetric:

But the following is not:

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

分析：

既然是镜面，那么直接翻转，然后查看翻转的树与原树是否一致。。我承认这个办法很low。。我先刷完这些题再说吧。。

代码：

/**
 * 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;
        
        TreeNode *mirror=new TreeNode(root->val);
        
        queue<TreeNode*> reTree;
        queue<TreeNode*> reTreeMirror;
        
        reTree.push(root);
        reTreeMirror.push(mirror);
        
        while(!reTree.empty()){
            TreeNode* tmp=reTree.front();
            reTree.pop();
            TreeNode* tmpMirror=reTreeMirror.front();
            reTreeMirror.pop();
            
            if(tmp->left){
                tmpMirror->right=new TreeNode((tmp->left)->val);
                reTree.push(tmp->left);
                reTreeMirror.push(tmpMirror->right);
            }
            if(tmp->right){
                tmpMirror->left=new TreeNode((tmp->right)->val);
                reTree.push(tmp->right);
                reTreeMirror.push(tmpMirror->left);
            }
        }
        
        
        queue<TreeNode*> cr;
        queue<TreeNode*> cm;
        
        cr.push(root);
        cm.push(mirror);
        
        while(!cr.empty()){
            TreeNode* tmpR=cr.front();
            TreeNode* tmpM=cm.front();
            cr.pop();
            cm.pop();
            
            if((tmpR->left==NULL&&tmpM->left!=NULL)||(tmpM->left==NULL&&tmpR->left!=NULL)||(tmpR->right==NULL&&tmpM->right!=NULL)||(tmpM->right==NULL&&tmpR->right!=NULL)||tmpR->val!=tmpM->val)return false;
            
            if(tmpR->left){
                cr.push(tmpR->left);
                cm.push(tmpM->left);
            }
            if(tmpR->right){
                cr.push(tmpR->right);
                cm.push(tmpM->right);
            }
        }
        
        return true;
    }
};