题目：
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? > read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/
2 3
/
4
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".题解：
判断左右子树是否对称。
解法一：递归的解法：
1 public boolean isSymmetricTree(TreeNode p,TreeNode q){ 2 if(p == null&&q == null) 3 return true; 4 if(p == null||q == null) 5 return false; 6 return (p.val == q.val) && isSymmetricTree(p.left, q.right) && isSymmetricTree(p.right, q.left); 7 } 8 9 public boolean isSymmetric(TreeNode root) { 10 if(root==null) 11 return true; 12 13 return isSymmetricTree(root.left,root.right); 14 }
解法二：非递归解法：
1 public boolean isSymmetric(TreeNode root) { 2 if(root == null) 3 return true; 4 if(root.left == null && root.right == null) 5 return true; 6 if(root.left == null || root.right == null) 7 return false; 8 LinkedList<TreeNode> q1 = new LinkedList<TreeNode>(); 9 LinkedList<TreeNode> q2 = new LinkedList<TreeNode>(); 10 q1.add(root.left); 11 q2.add(root.right); 12 while(!q1.isEmpty() && !q2.isEmpty()){ 13 TreeNode n1 = q1.poll(); 14 TreeNode n2 = q2.poll(); 15 16 if(n1.val != n2.val) 17 return false; 18 if((n1.left == null && n2.right != null) || (n1.left != null && n2.right == null)) 19 return false; 20 if((n1.right == null && n2.left != null) || (n1.right != null && n2.left == null)) 21 return false; 22 23 if(n1.left != null && n2.right != null){ 24 q1.add(n1.left); 25 q2.add(n2.right); 26 } 27 28 if(n1.right != null && n2.left != null){ 29 q1.add(n1.right); 30 q2.add(n2.left); 31 } 32 } 33 return true; 34 }
reference：
http://blog.csdn.net/linhuanmars/article/details/23072829