LeetCode - Easy - 572. Subtree of Another Tree-蒲公英云

LeetCode - Easy - 572. Subtree of Another Tree

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].
− 1 0 4 < = r o o t . v a l < = 1 0 4 -10^4 <= root.val <= 10^4 −104<=root.val<=104
− 1 0 4 < = s u b R o o t . v a l < = 1 0 4 -10^4 <= subRoot.val <= 10^4 −104<=subRoot.val<=104

Analysis

考查前序遍历。

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 && node2 != null) { 
            if(node1.val != node2.val) 
                return false;
            return equals(node1.left, node2.left) && //
                    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)));
    }
}