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
还是很简单的,递归方法如下:
/**
* Definition for binary tree
* 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;
}
TreeNode left = root.left;
TreeNode right = root.right;
return querySymmetric(left,right);
}
public boolean querySymmetric(TreeNode left, TreeNode right){
if(left == null && right == null){
return true;
}
if(left == null && right != null){
return false;
}
if(left != null && right == null){
return false;
}
if(left.val == right.val){
if(!querySymmetric(left.left,right.right)){
return false;
}
if(!querySymmetric(left.right,right.left)){
return false;
}
return true;
}else{
return false;
}
}
}