Symmetric Tree [LEETCODE]

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

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

 

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

 

Note:
Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ's Binary Tree Serialization:

The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.

Here's an example:

   1
  / \
 2   3
    /
   4
    \
     5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".

 

---

First I thought this problem could be solved by a pre-order traverse of the tree. Then I found I was wrong. The following tree will generate a list like "{2,3,1,2,3}", which is symmetric while the tree is not.

    1
   / \
  2   2
   \   \
   3    3

Then the serialization of tree enlightened me. The mark "#" could  be used to represent a NULL child. Why couldn't I use it too? So I got the following tree which generates a list as "{#,2,3,1,#,2,3}" when practice a pre-order tarverse.

Apparently the list is not symmetric any more.

    1
   / \
  2   2
 / \ / \
#  3 #  3

I use a vector<int *> to store pre-order tarverse of the tree, then I can store "#" as a NULL pointer. Here's the code:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <sstream>
class Solution {
public:
    bool isSymmetric(TreeNode *root) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        if(NULL == root) return true;
        vector<int*> vec = treeToVec(root);
        for(int i = 0; i < vec.size() - 1 - i; i++) {
            if(NULL == vec[i] && NULL == vec[vec.size()- 1 - i] || 
            NULL != vec[i] && NULL != vec[vec.size()- 1 - i] && *(vec[i]) == *(vec[vec.size()- 1 - i])) {
                continue;
            }
            return false;
        }
        return true;
        
    }
    vector<int*> treeToVec(TreeNode *root) {
        vector<int*> result;
        if(NULL == root) {
            result.push_back(NULL);
            return result;
        }
        
        vector<int*> left_vec = treeToVec(root->left);
        vector<int*> right_vec =  treeToVec(root->right);
        for(int i = 0; i < left_vec.size(); i++) {
            result.push_back(left_vec[i]);
        }
        result.push_back(&(root->val));
        for(int i = 0; i < right_vec.size(); i++) {
            result.push_back(right_vec[i]);
        }
        return result;
    }
};