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輸入兩棵二叉樹(shù)A，B，判斷B是不是A的子結(jié)構(gòu)。（我們約定空樹(shù)不是任意一個(gè)樹(shù)的子結(jié)構(gòu)） 假如給定A為{8,8,7,9,2,#,#,#,#,4,7}，B為{8,9,2}，2個(gè)樹(shù)的結(jié)構(gòu)如下，可以看出B是A的子結(jié)構(gòu) 數(shù)據(jù)范圍: 0 0
示例1
輸入
{8,8,7,9,2,#,#,#,#,4,7},{8,9,2}
輸出
true
示例2
輸入
{1,2,3,4,5},{2,4}
輸出
true
示例3
輸入
{1,2,3},{3,1}
輸出
false
加載中...
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ public boolean HasSubtree (TreeNode pRoot1, TreeNode pRoot2) { // write code here } }
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * }; */ class Solution { public: /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2) { // write code here } };
#coding:utf-8 # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 # # # @param pRoot1 TreeNode類 # @param pRoot2 TreeNode類 # @return bool布爾型 # class Solution: def HasSubtree(self , pRoot1 , pRoot2 ): # write code here
using System; using System.Collections.Generic; /* public class TreeNode { public int val; public TreeNode left; public TreeNode right; public TreeNode (int x) { val = x; } } */ class Solution { /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ public bool HasSubtree (TreeNode pRoot1, TreeNode pRoot2) { // write code here } }
/* * function TreeNode(x) { * this.val = x; * this.left = null; * this.right = null; * } */ /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ function HasSubtree( pRoot1 , pRoot2 ) { // write code here } module.exports = { HasSubtree : HasSubtree };
<?php /*class TreeNode{ var $val; var $left = NULL; var $right = NULL; function __construct($val){ $this->val = $val; } }*/ /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ function HasSubtree( $pRoot1 , $pRoot2 ) { // write code here }
# class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # # 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 # # # @param pRoot1 TreeNode類 # @param pRoot2 TreeNode類 # @return bool布爾型 # class Solution: def HasSubtree(self , pRoot1: TreeNode, pRoot2: TreeNode) -> bool: # write code here
package main import "fmt" import . "nc_tools" /* * type TreeNode struct { * Val int * Left *TreeNode * Right *TreeNode * } */ /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ func HasSubtree( pRoot1 *TreeNode , pRoot2 *TreeNode ) bool { // write code here }
/** * struct TreeNode { * int val; * struct TreeNode *left; * struct TreeNode *right; * }; */ /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ bool HasSubtree(struct TreeNode* pRoot1, struct TreeNode* pRoot2 ) { // write code here }
# class TreeNode # attr_accessor :val, :left, :right # def initialize(val, left = nil, right = nil) # @val, @left, @right = val, left, right # end # end # # 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 # # # @param pRoot1 TreeNode類 # @param pRoot2 TreeNode類 # @return bool布爾型 # class Solution def HasSubtree(pRoot1, pRoot2) # write code here end end
/** * class TreeNode(var val: Int) { * var left: TreeNode = null * var right: TreeNode = null * } */ object Solution { /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ def HasSubtree(pRoot1: TreeNode,pRoot2: TreeNode): Boolean = { // write code here } }
/** * class TreeNode(var `val`: Int) { * var left: TreeNode? = null * var right: TreeNode? = null * } */ object Solution { /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ fun HasSubtree(pRoot1: TreeNode?,pRoot2: TreeNode?): Boolean { // write code here } }
import java.util.*; /* * public class TreeNode { * int val = 0; * TreeNode left = null; * TreeNode right = null; * public TreeNode(int val) { * this.val = val; * } * } */ public class Solution { /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ public boolean HasSubtree (TreeNode pRoot1, TreeNode pRoot2) { // write code here } }
/*class TreeNode { * val: number * left: TreeNode | null * right: TreeNode | null * constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) { * this.val = (val===undefined ? 0 : val) * this.left = (left===undefined ? null : left) * this.right = (right===undefined ? null : right) * } * } */ /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ export function HasSubtree(pRoot1: TreeNode, pRoot2: TreeNode): boolean { // write code here }
/** * public class TreeNode { * public var val: Int * public var left: TreeNode? * public var right: TreeNode? * public init(_ val: Int=0, _ left: TreeNode?=nil, _ right: TreeNode?=nil) { * self.val = val * self.left = left * self.right = right * } */ public class Solution { /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ func HasSubtree ( _ pRoot1: TreeNode?, _ pRoot2: TreeNode?) -> Bool { // write code here } }
/** * #[derive(PartialEq, Eq, Debug, Clone)] * pub struct TreeNode { * pub val: i32, * pub left: Option<Box<TreeNode>>, * pub right: Option<Box<TreeNode>>, * } * * impl TreeNode { * #[inline] * fn new(val: i32) -> Self { * TreeNode { * val: val, * left: None, * right: None, * } * } * } */ struct Solution{ } impl Solution { fn new() -> Self { Solution{} } /** * 代碼中的類名、方法名、參數(shù)名已經(jīng)指定，請(qǐng)勿修改，直接返回方法規(guī)定的值即可 * * * @param pRoot1 TreeNode類 * @param pRoot2 TreeNode類 * @return bool布爾型 */ pub fn HasSubtree(&self, pRoot1: Option<Box<TreeNode>>, pRoot2: Option<Box<TreeNode>>) -> bool { // write code here } }
{8,8,7,9,2,#,#,#,#,4,7},{8,9,2}
true
{1,2,3,4,5},{2,4}
true
{1,2,3},{3,1}
false