【算法题】对称二叉树判断

解法1：

分层遍历二叉树，判断每一层的节点是否轴对称。

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#include <iostream>#include <vector>#include <algorithm>#include <numeric>using namespace std;#define debug_struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;    TreeNode(int x) :        val(x), left(NULL), right(NULL) {}};bool isok(vector<TreeNode*>& vec){    int i = 0, j = vec.size() - 1;    while (i < j)    {        if ((vec[i] == NULL&&vec[j] != NULL) || (vec[i] != NULL&&vec[j] == NULL))        {            return false;        }        if (vec[i]==NULL&&vec[j]==NULL)//考虑对称的空节点        {            i++;            j--;            continue;        }        if (vec[i]->val != vec[j]->val)        {            return false;        }        i++;        j--;    }    return true;}bool isSymmetrical(TreeNode* pRoot){        if (pRoot == NULL)        {            return true;        }        vector<TreeNode*> vec_old;        vector<TreeNode*> vec_new;        vec_old.push_back(pRoot);        while (!vec_old.empty())        {            if (!isok(vec_old))            {                return false;            }            for (auto i = 0; i < vec_old.size(); ++i)            {                if (vec_old[i] == NULL)                {                    continue;                }                vec_new.push_back(vec_old[i]->left);                vec_new.push_back(vec_old[i]->right);            }            swap(vec_old, vec_new);            vec_new.clear();        }        return true;}int p = 0;TreeNode* DeSerialByPre(char* str){    if (str[p] == '#'){        p += 2;        return NULL;    }    TreeNode* node = new TreeNode(str[p]);    p += 2;    node->left = DeSerialByPre(str);    node->right = DeSerialByPre(str);    return node;}int main(){    char* str = "8!6!5!#!#!7!#!#!6!7!#!#!5!#!#!";    TreeNode* root;    root = DeSerialByPre(str);    cout<<isSymmetrical(root);#ifdef debug_#else#endif    return 0;}

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解法2：前序遍历与对称前序遍历

bool isSymmetrical(TreeNode* pRoot1, TreeNode* pRoot2){    if (pRoot1 == NULL && pRoot2 == NULL)    {        return true;    }    if (pRoot1 == NULL || pRoot2 == NULL)    {        return false;    }    if (pRoot1->val != pRoot2->val)    {        return false;    }    return isSymmetrical(pRoot1->left, pRoot2->right) && isSymmetrical(pRoot1->right, pRoot2->left);}bool isSymmetrical(TreeNode* pRoot){    return isSymmetrical(pRoot, pRoot);}

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