Total Accepted: 93600
Total Submissions: 281557
Difficulty: Easy
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.
Java:
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23 * 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 node1 , TreeNode node2){
if(node1 == null && node2 == null) return true;
if(node1 == null || node2 == null) return false;
return node1.val == node2.val
&&isSymmetric(node1.left, node2.right)
&& isSymmetric(node1.right,node2.left);
}
}
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33/**
* 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;
Queue<TreeNode> q = new LinkedList<TreeNode>();
q.offer(root.left);
q.offer(root.right);
while (!q.isEmpty()) {
TreeNode l = q.poll();
TreeNode r = q.poll();
if (l == null && r == null)
continue;
if (l == null || r == null)
return false;
if(l.val!=r.val)
return false;
q.offer(l.left);
q.offer(r.right);
q.offer(l.right);
q.offer(r.left);
}
return true;
}
}
