Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3] is not:
1
/ \
2 2
\ \
3 3
Note:
Bonus points if you could solve it both recursively and iteratively.
判断一棵树是否是对称的。
/**
* Definition for a binary tree node.
* 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;
if( root.left == null && root.right == null)
return true;
if( root.left == null || root.right == null)
return false; return getResult(root.left,root.right); } public boolean getResult(TreeNode left,TreeNode right){
if( left == null && right == null)
return true;
if( left == null || right == null)
return false;
if( left.val != right.val )
return false;
if( getResult(left.left,right.right) )
return getResult(left.right,right.left);
return false; } }
也可以使用队列来解决这个问题。