标签:算法
问题描述:
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.
代码:
递归方法
public class Symmetric_Tree { //java
	public class TreeNode {
	      int val;
	      TreeNode left;
	      TreeNode right;
	      TreeNode(int x) { val = x; }
	 }
	//recursively
	public boolean isSymmetric(TreeNode root) {
      if(root == null)
    	  return true;
     
      return isSymmetricTree(root.left, root.right);
      
    }
	
	public boolean isSymmetricTree(TreeNode left, TreeNode right){
		if(left == null && right == null)
			return true;
		if(left == null || right == null ||left.val != right.val)
			return false;
		
		boolean lr = isSymmetricTree(left.left, right.right);
		boolean rl = isSymmetricTree(left.right, right.left);
		
		return lr && rl;
	}
}
标签:算法
原文地址:http://blog.csdn.net/chenlei0630/article/details/40514465