algorithms/symmetricTree/symmetricTree.cpp

// Source : https://oj.leetcode.com/problems/symmetric-tree/
// Author : Hao Chen
// Date   : 2014-06-28

/********************************************************************************** 
* 
* 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}". 
* 
*               
**********************************************************************************/

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    Solution(){
        srand(time(NULL));
    }

    bool isSymmetric(TreeNode *root) {
        if (root == NULL ) return true;
        return isSymmetric(root->left, root->right);
        
    }
    
    bool isSymmetric(TreeNode *p, TreeNode *q){
        if (random()%2){
           return isSymmetric1(p, q);
        }
        return isSymmetric2(p, q);
    }
    
    bool isSymmetric1(TreeNode *p, TreeNode *q){
        if (p==NULL && q==NULL) return true;
        if (p==NULL || q==NULL) return false;
        
        return (p->val == q->val) &&
                isSymmetric(p->left, q->right) &&
                isSymmetric(p->right, q->left);
    }
    
    bool isSymmetric2(TreeNode *p, TreeNode *q){

        queue<TreeNode*> q1;
        queue<TreeNode*> q2;
        q1.push(p);
        q2.push(q);
        while(q1.size()>0 && q2.size()>0){
            TreeNode* p1 = q1.front();
            q1.pop();
            TreeNode* p2 = q2.front();
            q2.pop();
            if (p1==NULL && p2==NULL) continue;
            if (p1==NULL || p2==NULL) return false;
            
            if (p1->val != p2->val) return false;
            
            q1.push(p1->left);
            q2.push(p2->right);

            q1.push(p1->right);
            q2.push(p2->left);

        }
        return true;
        
    }
    
};