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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 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?
OJ‘s Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where ‘#‘ signifies a path terminator where no node exists below.
Here‘s an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
1. 递归
2. iterative,建立两个queue, 一个从左往右，一个从右往左，镜像比较
仔细考虑每种情况，很容易漏掉。
Java code :
recursion:
public class Solution { public boolean isSymmetric(TreeNode root) { if(root == null) { return true; } return isSymmetric(root.left, root.right); } public boolean isSymmetric(TreeNode l, TreeNode r) { if(l == null && r == null){ return true; }else if(l == null || r == null) { return false; } if(l.val != r.val ){ return false; } if(!isSymmetric(l.left,r.right)) { return false; } if(!isSymmetric(l.right,r.left)) { return false; } return true; } }
iterative
public class Solution { public boolean isSymmetric(TreeNode root) { //use iterative if(root == null){ return true; } Queue<TreeNode> left = new LinkedList<TreeNode>(); Queue<TreeNode> right = new LinkedList<TreeNode>(); left.add(root.left); right.add(root.right); while(!left.isEmpty() && !right.isEmpty()) { TreeNode l = left.remove(); TreeNode r = right.remove(); if(l== null && r== null) { continue; }else if(l == null || r == null){ return false; } if(l.val != r.val) { return false; }else { left.add(l.left); left.add(l.right); right.add(r.right); right.add(r.left); } } return true; } }
Reference:
1. http://www.programcreek.com/2014/03/leetcode-symmetric-tree-java/
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原文地址：http://www.cnblogs.com/anne-vista/p/4812014.html
