技術標籤：LeetCode演算法與資料結構TreeDFSBFS

Topic

• Tree
• Depth-First Search
• Breadth-first Search

Description

https://leetcode.com/problems/symmetric-tree/

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

Follow up: Solve it both recursively and iteratively.

Analysis

方法一：BFS

方法二：DFS（遞迴）

Submission

import java.util.Arrays;
import java.util.LinkedList;

import com.lun.util.BinaryTree.TreeNode;

public class SymmetricTree
 
 {

	//方法一:我寫的BFS
	public boolean isSymmetric1(TreeNode root) {
		if (root == null) return true;
		LinkedList<TreeNode> queue = new LinkedList<>(Arrays.asList(root.left, root.right));

		while (!queue.isEmpty()) {
			int length = queue.size();

			for (int i = 0, j = length - 1; i < length /
 
 2; i++, j--) {
				TreeNode tn1 = queue.get(i);
				TreeNode tn2 = queue.get(j);

				if (tn1 == null && tn2 != null || tn1 != null && tn2 == null) {
					return false;
				}

				if (tn1 != null && tn2 != null) {
					if (tn1.val != tn2.val) {
						return false;
					}
				}
			}

			for(int i = length - 1; i >= 0; i--) {
				TreeNode node = queue.get(i);
				queue.remove(i);
				if(node != null)
					queue.addAll(i, Arrays.asList(node.left, node.right));
			}
		}
		return true;
	}

	//方法一:別人寫的BFS
	public boolean isSymmetric2(TreeNode root) {
		if (root == null) return true;
		LinkedList<TreeNode> q = new LinkedList<>(Arrays.asList(root.left, root.right));
		while (!q.isEmpty()) {
			TreeNode left = q.poll(), right = q.poll();
			if (left == null && right == null) continue;
			if (left == null || right == null || left.val != right.val) return false;
			q.addAll(Arrays.asList(left.left, right.right, left.right, right.left));
		}
		return true;
	}
	
	//方法二：
	public boolean isSymmetric3(TreeNode root) {
		return root == null || isSymmetric3(root, root);
	}
	
	private boolean isSymmetric3(TreeNode root1, TreeNode root2) {
		if(root1 == null && root2 != null || root1 != null && root2 == null)
			return false;
		
		if(root1 != null && root2 != null) {
			
			if(root1.val != root2.val) {
				return false;
			}
			return isSymmetric3(root1.left, root2.right) && isSymmetric3(root1.right, root2.left);
		}
		
		return true;
	}
}

Test

import static org.junit.Assert.*;

import org.junit.Before;
import org.junit.Test;

import com.lun.util.BinaryTree.TreeNode;

public class SymmetricTreeTest {

	private TreeNode root = null;
	
	@Before
	public void init() {
		root = new TreeNode(1);
		root.left = new TreeNode(2);
		root.right = new TreeNode(2);
	}
	
	
	@Test
	public void test1() {
		SymmetricTree obj = new SymmetricTree();

		root.left.left = new TreeNode(3);
		root.left.right = new TreeNode(4);
		root.right.left = new TreeNode(4);
		root.right.right = new TreeNode(3);
		
		assertTrue(obj.isSymmetric1(root));
		assertTrue(obj.isSymmetric2(root));
		assertTrue(obj.isSymmetric3(root));
	}
	
	@Test
	public void test2() {
		SymmetricTree obj = new SymmetricTree();
		
		root.left.right = new TreeNode(3);
		root.right.right = new TreeNode(3);
		
		assertFalse(obj.isSymmetric1(root));
		assertFalse(obj.isSymmetric2(root));
		assertFalse(obj.isSymmetric3(root));
	}
}