LeetCode - Easy - 572. Subtree of Another Tree

語言: CN / TW / HK

Topic

  • Tree

Description

https://leetcode.com/problems/subtree-of-another-tree/

Given the roots of two binary trees root and subRoot, return true if there is a subtree of root with the same structure and node values of subRoot and false otherwise.

A subtree of a binary tree tree is a tree that consists of a node in tree and all of this node's descendants. The tree tree could also be considered as a subtree of itself.

Example 1:

Input: root = [3,4,5,1,2], subRoot = [4,1,2]
Output: true

Example 2:

Input: root = [3,4,5,1,2,null,null,null,null,0], subRoot = [4,1,2]
Output: false

Constraints:

  • The number of nodes in the root tree is in the range [1, 2000].
  • The number of nodes in the subRoot tree is in the range [1, 1000].
  • $-10^4 <= root.val <= 10^4$
  • $-10^4 <= subRoot.val <= 10^4$

Analysis

考查前序遍歷。

Submission

import com.lun.util.BinaryTree.TreeNode;

public class SubtreeOfAnotherTree {
    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
    	if(root == null ^ subRoot == null)
    		return false;
    	
        if(equals(root, subRoot))
        	return true;

    	return isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);
    }
    
    public boolean equals(TreeNode node1, TreeNode node2) {
    	if(node1 == null ^ node2 == null)
    		return false;
    	
    	if(node1 != null &amp;&amp; node2 != null) {
    		if(node1.val != node2.val) 
    			return false;

    		return equals(node1.left, node2.left) &amp;&amp; //
    				equals(node1.right, node2.right);
    	}
    	return true;
    }
}

Test

import static org.junit.Assert.*;
import static com.lun.util.BinaryTree.*;
import org.junit.Test;


public class SubtreeOfAnotherTreeTest {

	@Test
	public void test() {
		SubtreeOfAnotherTree obj = new SubtreeOfAnotherTree();

		assertTrue(obj.isSubtree(integers2BinaryTree(3, 4, 5, 1, 2), //
				integers2BinaryTree(4, 1, 2)));
		
		assertFalse(obj.isSubtree(integers2BinaryTree(3, 4, 5, 1, 2, null, null, null, null, 0),// 
				integers2BinaryTree(4, 1, 2)));
	}
}