[leetcode] Symmetric Tree

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.

package com.wyt.leetcodeOJ;/** * @author wangyitao * @Date 2015-01-26 * @version 1.0 * @Description * 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? * */public class SymmetricTree {public static void main(String[] args) {TreeNode root = new TreeNode(1);TreeNode node1 = new TreeNode(2);TreeNode node2 = new TreeNode(2);TreeNode node3 = new TreeNode(3);TreeNode node4 = new TreeNode(4);TreeNode node5 = new TreeNode(4);TreeNode node6 = new TreeNode(3);root.left = node1;root.right = node2;node1.left = node3;node1.right = node4;node2.left = node5;node2.right = node6;System.out.println(isSymmetric(root));}public static boolean isSymmetric(TreeNode root) {if (root == null) {return true;} else {return isRLSymmetric(root.left, root.right);}}private static boolean isRLSymmetric (TreeNode left, TreeNode right) {if (left == null && right == null) {return true;} else {if (left == null || right == null) {return false;}}if (left.val != right.val) {return false;}if (!isEquals(left.left, right.right) || !isEquals(left.right, right.left)) {return false;}return isRLSymmetric(left.right, right.left) && isRLSymmetric(left.left, right.right);}private static boolean isEquals (TreeNode t1, TreeNode t2) {if (t1!= null && t2!=null) {return t1.val ==t2.val;} else {return t1 == t2;}}}

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