LeetCode OJ--Symmetric Tree **

http://oj.leetcode.com/problems/symmetric-tree/

判断树是否对称

class Solution {
public:
bool compare(TreeNode *left, TreeNode *right)
{
    if(left == NULL && right != NULL)
        return false;
    if(left != NULL && right == NULL)
        return false;
    if(left == NULL && right == NULL)
        return true;
    if(left->val != right->val)
        return false;
    return compare(left->left, right->right) && compare(left->right, right->left);
}
    bool isSymmetric(TreeNode *root) {
        if(root == NULL)
            return true;
        return compare(root->left, root->right);
    }
};

上面是卡哒的代码。

本来想的是中序遍历，然后判断是否回文。代码如下：

 struct TreeNode {
   int val;
   TreeNode *left;
   TreeNode *right;
   TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 };
 vector<char> ans;
void leftFirstTravel(TreeNode *root)
{
    if(root->left == NULL)
        ans.push_back('#');
    else 
        leftFirstTravel(root->left);

    ans.push_back(root->val);

    if(root->right == NULL)
        ans.push_back('#');
    else
        leftFirstTravel(root->right);
    return;
}

class Solution {
public:
    
    bool isSymmetric(TreeNode *root) {    
        if(root == NULL)
            return true;
        ans.clear();   
        leftFirstTravel(root);

        int size = ans.size();
        int mid = size/2;
        for(int i = 1;mid - i>= 0;i++)
        {
            if(ans[mid-i] != ans[mid+i] )
                return false;
        }
        return true;
    }
};

Input:	{1,2,3,3,#,2,#}
Output:	true
Expected:	false

但是这样遍历出来的是 3 2 1 2 3 即便是加了#，是# 3 # 2 # 1 # 2 # 3 # ，但实际上不是。嗯嗯，要更加深刻的理解递归。