2019-05-27

101. Symmetric Tree
     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.

/**
 * 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) {
        
        queue<TreeNode*> q;
        q.push(root);
        q.push(root);
        
        while(!q.empty()){
            TreeNode* t1 = q.front();
            q.pop();
            TreeNode* t2 = q.front();
            q.pop();
            
            if(t1 == nullptr && t2 == nullptr) continue;
            if(t1 == nullptr || t2 == nullptr) return false;
            if(t1->val != t2->val) return false;
            
            q.push(t1->left);
            q.push(t2->right);
            q.push(t1->right);
            q.push(t2->left);
        }
        return true;
    }
};

/**
 * 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) {
        return is_mirrow(root, root);
    }
    
    bool is_mirrow(TreeNode* t1, TreeNode* t2){
        if(t1 == nullptr && t2 == nullptr) return true;
        if(t1 == nullptr || t2 == nullptr) return false;
        return (t1->val == t2->val) && is_mirrow(t1->left, t2->right) && is_mirrow(t1->right, t2->left);
    }
};