【待解决】[LeetCode-101]-Symmetric Tree(判断两颗二叉树是否对称)

0. 题目相关

【题目解读】
给定两颗二叉树，对这两颗二叉树进行比较，判断这两棵二叉树是否对称

【原题描述】原题链接
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:

   1
  / \
 2   2
/ \ / \
3  4 4  3

But the following [1,2,2,null,3,null,3] is not:

   1
  / \
 2   2
  \   \
  3    3

Note:
Bonus points if you could solve it both recursively and iteratively.
【难度】Easy

1. Solution

自己用的中根序遍历根节点的左右子树，然后进行比较。发现这种方法对于例子中**[1,2,2,null,3,null,3]**这种情况就不能正确判断，还需要进行修改。

有问题的代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#define Max_Size 100
#include <stack>
class Solution {
private:
    //由于有ret这个数组，不能使用递归版本
    int inOrder(TreeNode* root, int ret[])
    {
        int index = 0;
        stack<TreeNode*> Tstack;
        TreeNode* pre = root;
        while(pre != NULL && !Tstack.empty())
        {
            if(pre != NULL)
            {
                Tstack.push(pre);
                pre = pre->left;
            }else
            {
                pre = Tstack.top();
                ret[index++] = pre->val;
                pre = pre->right;     
                Tstack.pop();
            }
        }
        
        return index;
    }
    
public:
    bool isSymmetric(TreeNode* root) {
        if(root == NULL)
            return false;
        
        int lTree[Max_Size] = {0};
        int rTree[Max_Size] = {0};
        
        TreeNode *left = root->left;
        TreeNode *right = root->right;
        
        int lLen = inOrder(left, lTree);
        int rLen = inOrder(right, rTree);
        
        if(lLen != rLen)
            return false;

        for(int index = 0; index < lLen; index ++)
        {
            if(lTree[index] != rTree[rLen-index])
                return false;
        }
        
        return true;
    }
};