101. Symmetric Tree-优快云博客

本文链接：https://blog.youkuaiyun.com/csg3140100993/article/details/103119157

本文介绍了一种检查二叉树是否对称的方法，通过递归和迭代两种方式实现，利用前序遍历和对称遍历比较左右子树是否镜像对称。

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101. Symmetric Tree

Easy

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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.

/**
 * 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;
        vector<int> lftree;
        vector<int> rgtree;
        tra1(root->left,lftree);
        tra2(root->right,rgtree);
        //prin(lftree);
        //prin(rgtree);
        if(lftree!=rgtree) return false;
        else return true;
    }
    void tra1(TreeNode* root,vector<int>& vt){//前序遍历左子树
        if(root==NULL){
            vt.push_back(0);
            return;
        }
        tra1(root->left,vt);
        tra1(root->right,vt);
        vt.push_back(root->val);
    }
    void tra2(TreeNode* root,vector<int>& vt){//对称的遍历右子树
        if(root==NULL){
            vt.push_back(0);
            return;
        }
        tra2(root->right,vt);
        tra2(root->left,vt);
        vt.push_back(root->val);
    }
    /*void prin(vector<int> a){
        for(int i=0;i<a.size();i++){
            cout<<a[i]<<' ';
        }
        cout<<endl;
    }*/
};