leetcode101

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */
public classSolution {
    Queue<TreeNode> Q = new Queue<TreeNode>();
        private List<string>GetFloorString()
        {
            var floor = new List<string>();
            var count = 0;
            while (Q.Count > 0)
            {
                var n =Q.Dequeue();
                if (n.left != null)
                {
                    floor.Add(n.left.val.ToString());
                    Q.Enqueue(n.left);
                    count++;
                }
                else
                {
                    floor.Add("x");
                }
                if (n.right != null)
                {
                    floor.Add(n.right.val.ToString());
                    Q.Enqueue(n.right);
                    count++;
                }
                else
                {
                    floor.Add("x");
                }
            }
            returnfloor;
        }
        public boolIsSymmetric(TreeNode root)
        {
            if (root == null)
            {
                return true;
            }
            else
            {
                Q.Enqueue(root);
                var str =GetFloorString();
                int index = 0;//起始索引
                var xcount = 0;
                //len长度
                for (int len = 2; index + len <= str.Count && len > 0; len = 2 * (len -xcount))
                {
                    xcount = 0;
                    var l = new List<string>();//str.Substring(index, len);
                    for (int i = index; i < index + len; i++)
                    {
                        l.Add(str[i]);
                    }
                    var halflen = len / 2;
                    var l1 = new List<string>();
                    for (int i = 0; i < halflen; i++)
                    {
                        l1.Add(l[i]);
                    }
                    //var s2 = s.Substring(halflen, halflen);
                    var l2 = new List<string>();
                    for (int i = halflen; i < halflen + halflen; i++)
                    {
                        l2.Add(l[i]);
                    }
                    l2.Reverse();
                    var s1 = "";
                    var s2 = "";
                    for (int i = 0; i < l1.Count; i++)
                    {
                        s1 +=l1[i];
                    }
                    for (int i = 0; i < l2.Count; i++)
                    {
                        s2 +=l2[i];
                    }
                    if (s1 !=s2)
                    {
                        return false;
                    }
                    foreach (var c inl)
                    {
                        if (c == "x")
                        {
                            xcount++;
                        }
                    }
                    index = index +len;
                }
                return true;
            }
        }
}

1 classSolution:
2     def isSymmetric(self, root: 'TreeNode') -> 'bool':
3         if root ==None:
4             returnTrue
5         returnself.isSymmetric2(root.left,root.right)
6         
7 def isSymmetric2(self,left,right):
8         if left == None and right ==None:
9             returnTrue
10         if left == None or right ==None:
11             returnFalse
12         if left.val !=right.val:
13             returnFalse
14         return self.isSymmetric2(left.left,right.right) and self.isSymmetric2(left.right,right.left)