中国做网站找谁/种子在线资源搜索神器

给定一个二叉树，检查它是否是镜像对称的。

例如，二叉树 [1,2,2,3,4,4,3] 是对称的。

    1/ \2   2/ \ / \
3  4 4  3

但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:

    1/ \2   2\   \3    3

进阶：

你可以运用递归和迭代两种方法解决这个问题吗？

迭代：（广度优先遍历）

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode() {}*     TreeNode(int val) { this.val = val; }*     TreeNode(int val, TreeNode left, TreeNode right) {*         this.val = val;*         this.left = left;*         this.right = right;*     }* }*/
class Solution {public boolean isSymmetric(TreeNode root) {if(root==null||(root.left==null&&root.right==null)){return true;}//使用队列存储数据LinkedList<TreeNode> Queue=new LinkedList<TreeNode>();//先存储根的左右结点Queue.add(root.left);Queue.add(root.right);//while(Queue.size()>0){//从队列中取出对应得左右结点，进行比较TreeNode left= Queue.removeFirst();TreeNode right=Queue.removeFirst();  if(left==null && right==null){continue;}if(left==null||right==null){return false;}if(left.val!=right.val){return false;}//将此时左节点的左孩子，右节点的右孩子进行存储Queue.add(left.left);Queue.add(right.right);//将此时左节点的右孩子，右节点的左孩子进行存储Queue.add(left.right);Queue.add(right.left);}return true;}}

递归（深度优先遍历）

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