LeetCode-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.

使用队列，两两加入队列，两两出队列

public boolean isSymmetric(TreeNode root) {//树 my
        if(null==root){
            return true;
        }
        LinkedList<TreeNode> queue = new LinkedList<>();
        queue.add(root.left);
        queue.add(root.right);
        while(queue.size()!=0){
            TreeNode n1 = queue.poll();
            TreeNode n2 = queue.poll();
            if(n1==null&&null==n2) continue;
            if(null==n1||null==n2) return false;
            if(n1.val!=n2.val) return false;
            queue.add(n1.left);
            queue.add(n2.right);
            queue.add(n1.right);
            queue.add(n2.left);
        }
        return true;
    }

递归

class Solution {
    public boolean isSymmetric(TreeNode root) {//树 my
        if(null==root){
            return true;
        }
        return help(root.left,root.right);
    }
    private boolean help(TreeNode n1 ,TreeNode n2){
        if(n1==null&&null==n2) return true;
        if(null==n1||null==n2) return false;
        if(n1.val!=n2.val) return false;
        return help(n1.left,n2.right)&&help(n1.right,n2.left);
    }
}