LeetCode_101. Symmetric Tree

 

101. Symmetric Tree

Easy

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.

 

package leetcode.easy;

import java.util.LinkedList;
import java.util.Queue;

public class SymmetricTree {
	@org.junit.Test
	public void test1() {
		TreeNode tn11 = new TreeNode(1);
		TreeNode tn21 = new TreeNode(2);
		TreeNode tn22 = new TreeNode(2);
		TreeNode tn31 = new TreeNode(3);
		TreeNode tn32 = new TreeNode(4);
		TreeNode tn33 = new TreeNode(4);
		TreeNode tn34 = new TreeNode(3);
		tn11.left = tn21;
		tn11.right = tn22;
		tn21.left = tn31;
		tn21.right = tn32;
		tn22.left = tn33;
		tn22.right = tn34;
		tn31.left = null;
		tn31.right = null;
		tn32.left = null;
		tn32.right = null;
		tn33.left = null;
		tn33.right = null;
		tn34.left = null;
		tn34.right = null;
		System.out.println(isSymmetric1(tn11));
		System.out.println(isSymmetric2(tn11));
	}

	@org.junit.Test
	public void test2() {
		TreeNode tn11 = new TreeNode(1);
		TreeNode tn21 = new TreeNode(2);
		TreeNode tn22 = new TreeNode(2);
		TreeNode tn32 = new TreeNode(4);
		TreeNode tn34 = new TreeNode(3);
		tn11.left = tn21;
		tn11.right = tn22;
		tn21.left = null;
		tn21.right = tn32;
		tn22.left = null;
		tn22.right = tn34;
		tn32.left = null;
		tn32.right = null;
		tn34.left = null;
		tn34.right = null;
		System.out.println(isSymmetric1(tn11));
		System.out.println(isSymmetric2(tn11));
	}

	public boolean isSymmetric1(TreeNode root) {
		return isMirror(root, root);
	}

	public boolean isMirror(TreeNode t1, TreeNode t2) {
		if (t1 == null && t2 == null) {
			return true;
		}
		if (t1 == null || t2 == null) {
			return false;
		}
		return (t1.val == t2.val) && isMirror(t1.right, t2.left) && isMirror(t1.left, t2.right);
	}

	public boolean isSymmetric2(TreeNode root) {
		Queue q = new LinkedList<>();
		q.add(root);
		q.add(root);
		while (!q.isEmpty()) {
			TreeNode t1 = q.poll();
			TreeNode t2 = q.poll();
			if (t1 == null && t2 == null) {
				continue;
			}
			if (t1 == null || t2 == null) {
				return false;
			}
			if (t1.val != t2.val) {
				return false;
			}
			q.add(t1.left);
			q.add(t2.right);
			q.add(t1.right);
			q.add(t2.left);
		}
		return true;
	}
}