Symmetric Tree @LeetCode

PROBLEM :

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
 
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SIMPLE c++ IMPLEMENTATION :(Recursive Solution)
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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

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

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SIMPLE c++ IMPLEMENTATION :(Iterative Solution - using Stack STL)
--------------------------------------------------------------------------------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root==NULL)
            return true ;
         
        stack<TreeNode*> a ;
        stack<TreeNode*> b ;
        a.push(root) ;
        b.push(root) ;
     
        while(!a.empty()&&!b.empty()){
            TreeNode* x ;
            TreeNode* y ;
         
            x=a.top() ;
            y=b.top() ;
         
            a.pop() ;
            b.pop() ;
         
            if(x==NULL&&y==NULL)
                continue ;
            if(x==NULL||y==NULL)
                return false ;
            if(x->val!=y->val)
                return false ;
         
            a.push(x->left) ;
            a.push(x->right) ;
         
            b.push(y->right) ;
            b.push(y->left) ;
        }
        return true ;
    }
};

--------------------------------------------------------------------------------
SIMPLE c++ IMPLEMENTATION :(Iterative Solution - using Queue STL)
--------------------------------------------------------------------------------

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */

class Solution {
public:
    bool isSymmetric(TreeNode* root) {
        if(root==NULL)
            return true ;
         
        queue<TreeNode*> a ;
        queue<TreeNode*> b ;
        a.push(root) ;
        b.push(root) ;
     
        while(!a.empty()&&!b.empty()){
            TreeNode* x ;
            TreeNode* y ;
         
            x=a.front() ;
            y=b.front() ;
         
            a.pop() ;
            b.pop() ;
         
            if(x==NULL&&y==NULL)
                continue ;
            if(x==NULL||y==NULL)
                return false ;
            if(x->val!=y->val)
                return false ;
         
            a.push(x->left) ;
            a.push(x->right) ;
         
            b.push(y->right) ;
            b.push(y->left) ;
        }
        return true ;
    }
};

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