LeetCode 101. Symmetric Tree

最新推荐文章于 2024-01-17 14:31:46 发布

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本文探讨了如何检查一个二叉树是否为自身镜像（即围绕其中心对称）。提供两种方法：一种通过递归方式遍历整棵树并进行比较；另一种采用迭代方法，利用队列逐层对比节点。

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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) {
        vector<int>a;
        vector<int>b;
        if(root==NULL)return true;
        else{
            if(root->left!=NULL&&root->right!=NULL){
            a=solvel(root->left,a);
            b=solver(root->right,b);

            }
            else if(root->left==NULL&&root->right==NULL)return true;
            else return false;
        }
        return a==b;
    }
    vector<int> solvel(TreeNode *root,vector<int>r){
        r.push_back(root->val);
        if(root->left!=NULL||root->right!=NULL){

            if(root->left!=NULL&&root->right==NULL){r.push_back(NULL);r=solvel(root->left,r);}
            if(root->left==NULL&&root->right!=NULL){r=solvel(root->right,r);r.push_back(NULL);}
            if(root->left!=NULL&&root->right!=NULL){
                r=solvel(root->left,r);
                r=solvel(root->right,r);

        }
        }

        return r;
    }
        vector<int> solver(TreeNode *root,vector<int>r){
        r.push_back(root->val);
        if(root->left!=NULL||root->right!=NULL){

            if(root->left!=NULL&&root->right==NULL){r=solver(root->left,r);r.push_back(NULL);}
            if(root->left==NULL&&root->right!=NULL){r.push_back(NULL);r=solver(root->right,r);}
            if(root->left!=NULL&&root->right!=NULL){
                r=solver(root->right,r);
                r=solver(root->left,r);
        }
        }


        return r;
    }
};

这是大神们写的：

class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        TreeNode *left, *right;
        if (!root)
            return true;

        queue<TreeNode*> q1, q2;
        q1.push(root->left);
        q2.push(root->right);
        while (!q1.empty() && !q2.empty()){
            left = q1.front();
            q1.pop();
            right = q2.front();
            q2.pop();
            if (NULL == left && NULL == right)
                continue;
            if (NULL == left || NULL == right)
                return false;
            if (left->val != right->val)
                return false;
            q1.push(left->left);
            q1.push(left->right);
            q2.push(right->right);
            q2.push(right->left);
        }
        return true;
    }
};