Symmetric Tree

错误，应该写一个help函数，接受左右两个参数

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.

recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        
        
        if(root ==null)
            return true;
        
        
        return help(root.left, root.right);
        
    }
    
    public boolean help(TreeNode left, TreeNode right){
        if(left == null && right == null){
            return true;
        }
        
        if((left == null && right != null ) || (left != null && right == null) || (left.val != right.val))
            return false;
        
        return help(left.left, right.right) && help(left.right, right.left);
    }
}

iterative:一个q就够了，注意判断一下queue里弹出的node是不是null，是null 的话 继续，如果不判断 下一步node left right时会报指针异常

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean isSymmetric(TreeNode root) {
        
        
        if(root ==null)
            return true;
        
        
        Queue<TreeNode> q1;
        
        q1 = new LinkedList<>();

        
        q1.offer(root.left);
        q1.offer(root.right);
        
        while(!q1.isEmpty()){
            TreeNode node1 = q1.poll();
            TreeNode node2 = q1.poll();
            if(node1 == null && node2 == null)
                continue;
            if((node1 == null && node2 != null) || (node1 !=null && node2 == null) || (node1.val != node2.val)){
                return false;
            }
            
            q1.offer(node1.left);
            q1.offer(node2.right);
            q1.offer(node1.right);
            q1.offer(node2.left);

            
        }
        
        return true;
        
    }
    
   
}