Leetcode-symmetric-tree

题目描述

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.

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}".

题目意思比较明确，判断一棵树是不是镜像树，从中间对称。

很明显运用递归的方法。

对于一个根节点root，需要判断root.left和root.right是否对称，同时需要知道root.left子树和root.right子树本身也是对称的，也就是root.left.right.val == root.right.left.val && root.left.left.val == root.right.right.val，如果是true，结合递归的思想即可判断出来整棵树的对称的。

 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean isSymmetric(TreeNode root) {

    	if(root == null)
    		return true;
    	return isSysmmetric(root.left, root.right);
    }
    
    public boolean isSysmmetric(TreeNode node1, TreeNode node2){
    	if(node1 == null && node2==null)
    		return true;
    	if(node1!=null && node2!=null){
    		if(node1.val == node2.val)
    			return isSysmmetric(node1.left,node2.right)&&isSysmmetric(
    					node1.right, node2.left);
    		else
    			return false;
    	}else{
    		return false;
    	}
    }
}