[leetcode] 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.

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

思路1：递归判断，注意null的处理。

思路2：迭代写法，有空一起整理。

public class Solution {

    public boolean isSymmetric(TreeNode root) {
        if (root == null)
            return true;
        return check(root.left, root.right);
    }

    private boolean check(TreeNode left, TreeNode right) {
        if (left == null && right == null)
            return true;
        if (left == null || right == null)
            return false;
        if (left.val != right.val)
            return false;
        return check(left.right, right.left) && check(left.left, right.right);

    }

    public static void main(String[] args) {
        TreeNode root = new TreeNode(10);
        root.left = new TreeNode(5);
        root.left.right = new TreeNode(9);
        root.right = new TreeNode(5);
        root.right.left = new TreeNode(9);

        System.out.println(new Solution().isSymmetric(root));

    }

}

第二遍记录：null check can be easier

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root==null){
            return true;
        }
        return isSym(root.left, root.right);
    }
    
    private boolean isSym(TreeNode left, TreeNode right){
        if(left == null || right == null){
            return left == right;
        }
        
        return left.val==right.val && isSym(left.left, right.right) && isSym(left.right, right.left);
        
    }
    
}

参考：

http://www.cnblogs.com/remlostime/archive/2012/11/15/2772230.html