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 class Solution {
    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");
                }
            }
            return floor;
        }

        public bool IsSymmetric(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 in l)
                    {
                        if (c == "x")
                        {
                            xcount++;
                        }
                    }

                    index = index + len;
                }

                return true;
            }
        }
}

https://leetcode.com/problems/symmetric-tree/#/description

 

补充一个python的实现：

class Solution:
    def isSymmetric(self, root: 'TreeNode') -> 'bool':
        if root == None:
            return True
        return self.isSymmetric2(root.left,root.right)
        
    def isSymmetric2(self,left,right):
        if left == None and right == None:
            return True
        if left == None or right == None:
            return False
        if left.val != right.val:
            return False
        return self.isSymmetric2(left.left,right.right) and self.isSymmetric2(left.right,right.left)

 

Java版本：

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null){
            return true;
        }else{
            return sameTree(root.left,root.right);
        }
    }

    public boolean sameTree(TreeNode left,TreeNode right){
        if(left == null && right == null){
            return true;
        }else if(left != null && right == null){
            return false;
        }else if(left == null && right != null){
            return false;
        }else{
            if(left.val != right.val){
                return false;
            }else{
                return sameTree(left.left,right.right) && sameTree(left.right,right.left);
            }
        }
    }
}