[leetcode刷题系列]Symmetric Tree

- - 递归版本先

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    bool isSy(TreeNode * left, TreeNode * right){
        if(left == 0 && right == 0)
            return true;
        if(left == 0 || right == 0)
            return false;
        if(left->val != right->val)
            return false;
        return isSy(left->left, right->right) && isSy(left->right, right->left);
    }
public:
    bool isSymmetric(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == 0)
            return true;
        if(root->left == 0 && root->right == 0)
            return true;
        return isSy(root->left, root->right);
    }
};

非递归版本， 通过判断每一层是否回文来判断

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
    bool ispl(vector<TreeNode* > & vc){
        int left = 0, right = vc.size() - 1;
        while(left < right){
            if(vc[left] == 0 && vc[right] == 0)
                ;
            else if(vc[left] == 0 || vc[right] == 0)
                return false;
            else if(vc[left]->val != vc[right]->val)
                return false;
            ++ left;
            -- right;
        }
        return true;
    }
public:
    bool isSymmetric(TreeNode *root) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if(root == 0)
            return true;
        int pre = 0, next = 1;
        vector<TreeNode* > vc[2];
        vc[0].push_back(root);
        while(vc[pre].size() > 0){
            for(int i = 0; i < vc[pre].size(); ++ i){
                TreeNode * now = vc[pre][i];
                if(now != 0){
                    vc[next].push_back(now->left);
                    vc[next].push_back(now->right);
                }
            }
            vc[pre].clear();
            if(!ispl(vc[next]))
                return false;
            swap(pre, next);
        }
        return true;
    }
};