[LeetCode]101. Symmetric Tree
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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.
这道题的思路其实和【100 sme tree】是比较类似的。
不过在这道题里面,递归的思路更容易写。
/** * 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; else return symme(root.left, root.right); } public boolean symme(TreeNode p, TreeNode q){ if(p==null && q==null) return true; if(p==null && q!=null) return false; if(p!=null && q==null) return false; if(p.val!=q.val) return false; else return(symme(p.left, q.right) && symme(p.right, q.left)); }}
我想了一下(没有去写),用queue来进行层序遍历也是可以的。
左子树的子节点的入队按照从右到左的顺序,右子树的子节点的入队按照从左到右的顺序,这样来依次检查节点是否一致。
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