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
一刷
题解：用recursion做思路很自然。
/**
* 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;
return isSymmetric(root.left, root.right);
}
public boolean isSymmetric(TreeNode left, TreeNode right){
if(left == null && left == right) return true;
if((left!=null && right == null) ||(left==null && right != null)) return false;
if(left.val != right.val) return false;
return isSymmetric(left.left, right.right)
&& isSymmetric(left.right, right.left);
}
}
二刷
recursion
/**
* 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;
return isSymmetric(root.left, root.right);
}
private boolean isSymmetric(TreeNode left, TreeNode right){
if(left == null && right == null) return true;
if(left == null || right == null) return false;
if(left.val == right.val){
return isSymmetric(left.left, right.right) &&
isSymmetric(left.right, right.left);
}
else return false;
}
}
