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풀어본 Algorithm 문제 정리

[OJ.leetcode] symmetric-tree

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


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.

**

 * Definition for binary tree

 * struct TreeNode {

 *     int val;

 *     TreeNode *left;

 *     TreeNode *right;

 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}

 * };

 */

 

 /*

  recursive call로 각 노드를 비교해야되는데 어떻게 비교해야할지가 관건

  left->left 와 right->right,

  left->right와 right->left를 매 recursive call 마다 비교해야하는 것을 알아차리는게 관건


  Time complexity: O(n)

 */


class Solution {

    public:

    bool isSymmetric(TreeNode *root) {

        if(root == NULL) return true;

        

        if(root->left && root->right) {

            return isSymmetric(root->left, root->right);

        } else if( root->left == NULL && root->right == NULL){

            return true;

        } else {

            return false;    

        }

    }

    

    bool isSymmetric(TreeNode* left, TreeNode* right) {

        if(left == NULL && right == NULL) return true;

        else if(left == NULL || right == NULL) return false;

        else if(left->val != right->val) return false;

        return isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left);

    }

};