leetcode_101_Symmetric Tree

描述：

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.

   1
  /  2   3
    /
   4
         5

代码：

class TreeNode{
		int val;
		TreeNode left;
		TreeNode right;
		TreeNode(int x){val=x;}
	}
	public boolean isSymmetric(TreeNode root) {
		if(root==null)
			return true;
        if(treeIsSymmetric(root.left, root.right))
        	return true;
		return false;
    }
	public boolean treeIsSymmetric(TreeNode p,TreeNode q)
	{
		if(p==null&&q==null)//左右子树都为空时对称
			return true;
		if(p==null||q==null)//左右子树有一个不为空时不对称
			return false;
		//子树结点值相等，左孩子的结点值等于右孩子的结点值。。。
		if(p.val==q.val&&treeIsSymmetric(p.left, q.right)&&treeIsSymmetric(p.right, q.left))
			return true;
		return false;
	}

结果：

leetcode_101_Symmetric Tree

原文：http://blog.csdn.net/mnmlist/article/details/44591291