Symmetric Tree

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree is symmetric:

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

 

But the following is not:

    1
   / \
  2   2
   \   \
   3    3

 

Note:
Bonus points if you could solve it both recursively and iteratively.

confused what "{1,#,2,3}" means?

思想：递推

java代码：

1. public boolean isSymmetric(TreeNode left,TreeNode right) {
2. if(left==null && right==null) return true;
3. if(left == null || right == null) return false;
4. if(left.val != right.val) return false;
5. boolean flag = isSymmetric(left.left,right.right) && isSymmetric(left.right,right.left);
6. return flag;
7. }
8. public boolean isSymmetric(TreeNode root) {
9. if(root==null) return true;
10. if(root.left==null && root.right==null) return true;
11. return isSymmetric(root.left,root.right);
12. }

思想二：迭代

注意事项：java 中Queue是一个接口，需要用LinkedList来初始化。提供来offer和poll来添加和移除数据，如果溢出，返回false或者不存在，则返回null。相对于add和remove安全一些，或者说add和remove会首先根据offer和poll的结果来判断是否存在异常。

Queue中的peek返回头节点，如果不存在则返回null。

1. public boolean isSymmetric(TreeNode root) {
2. if(root==null) return true;
3. if(root.left==null && root.right==null) return true;
4. Queue<TreeNode> q = new LinkedList<TreeNode>();
5. Queue<TreeNode> p = new LinkedList<TreeNode>();
6. if(root.left!=null) q.offer(root.left);
7. if(root.right!=null) p.offer(root.right);
8. while(q.peek() != null || p.peek() != null) {
9. TreeNode left = q.poll();
10. TreeNode right = p.poll();
11. if(left==null || right==null) return false;
12. if(left.val == right.val) {
13. if(left.left!=null && right.right!=null) {
14. q.offer(left.left);
15. p.offer(right.right);
16. } else if(left.left!=null || right.right!=null) {
17. return false;
18. }
19. if(left.right!=null && right.left!=null) {
20. q.offer(left.right);
21. p.offer(right.left);
22. } else if(left.right!=null || right.left!=null) {
23. return false;
24. }
25. } else {
26. return false;
27. }
28. }
29. return true;
30. }