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.
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/** * 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) { boolean
isSys = true; Deque<TreeNode> nodes = new
LinkedList<TreeNode>(); if(root != null){ nodes.addFirst(root); nodes.addLast(root); while(nodes.peek() != null){ Deque<TreeNode> nextLevel = new
LinkedList<TreeNode>(); while(nodes.peek() != null){ TreeNode first = nodes.pollFirst(); TreeNode last = nodes.pollLast(); if(first.val != last.val){ isSys = false; break; } else{ if((first.left != null
&& last.right == null) || (first.left == null
&& last.right != null)){ isSys = false; break; } if(first.left != null
&& last.right != null){ nextLevel.addFirst(first.left); nextLevel.addLast(last.right); } if((first.right != null
&& last.left == null) || (first.right == null
&& last.left != null)){ isSys = false; break; } if(first.right != null
&& last.left != null){ nextLevel.addFirst(first.right); nextLevel.addLast(last.left); } } } if(!isSys) break; else nodes = nextLevel; } } return
isSys; }} |
原文：http://www.cnblogs.com/averillzheng/p/3545281.html