目录
- [1- 思路](#1- 思路)
- [2- 实现](#2- 实现)
-
- [⭐437. 路径总和 III------题解思路](#⭐437. 路径总和 III——题解思路)
- [3- ACM 实现](#3- ACM 实现)
- 题目连接:437. 路径总和 III
1- 思路
前缀和+哈希表+dfs
① 前缀和
- 求二叉树的前缀和,每求一次用一个
sum
传参记录更新
② 哈希表
key
为前缀和 ,value
为出现频率- 用来记录前缀和出现的次数,原理就是如果
sum - targetSum
在前缀和的哈希中,则证明有目标和为targetSum
的路径。出现次数就是该哈希出现的次数
③ dfs递归三部
- 3.1 参数返回值,返回
void
,参数为sum
和root
- 3.2 终止条件,遇到
null
直接返回 - 3.3 递归处理:递归求前缀和,如果哈希表存在
sum - targetSum
且出现次数大于等于 1 ,收集res
- 更新
sum
- 递归 左节点
- 递归 右结点
- 回溯 更新
sum
- 更新
2- 实现
⭐437. 路径总和 III------题解思路
java
class Solution {
int targetSum;
int res = 0;
HashMap<Long,Integer> hash = new HashMap<>();
public int pathSum(TreeNode root, int targetSum) {
this.targetSum = targetSum;
hash.put(0L,1);
dfs(0L,root);
return res;
}
public void dfs(Long sum,TreeNode root){
// 终止
if(root==null){
return;
}
// 递归
sum+=root.val;
if(hash.containsKey(sum-targetSum) && hash.get(sum-targetSum)>=1){
res += hash.get(sum-targetSum);
}
hash.put((Long)sum,hash.getOrDefault(sum,0)+1);
dfs(sum,root.left);
dfs(sum,root.right);
hash.put((Long)sum,hash.get(sum)-1);
}
}
3- ACM 实现
java
package Daily_LC.Month8_Week4.Day138;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;
/**
* pathSum
*
* @author alcohol
* @Description
* @since 2024-08-24 13:40
*/
public class pathSum {
public static class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode() {
}
TreeNode(int val) {
this.val = val;
}
TreeNode(int val, TreeNode left, TreeNode right) {
this.val = val;
this.left = left;
this.right = right;
}
}
public static TreeNode build(String str) {
if (str == null || str.length() == 0) {
return null;
}
String input = str.replace("[", "");
input = input.replace("]", "");
String[] parts = input.split(",");
Integer[] nums = new Integer[parts.length];
for (int i = 0; i < parts.length; i++) {
if (!parts[i].equals("null")) {
nums[i] = Integer.parseInt(parts[i]);
} else {
nums[i] = null;
}
}
Queue<TreeNode> queue = new LinkedList<>();
TreeNode root = new TreeNode(nums[0]);
queue.offer(root);
int index = 1;
while (!queue.isEmpty() && index < parts.length) {
TreeNode node = queue.poll();
if (index < nums.length && nums[index] != null) {
node.left = new TreeNode(nums[index]);
queue.offer(node.left);
}
index++;
if (index < nums.length && nums[index] != null) {
node.right = new TreeNode(nums[index]);
queue.offer(node.right);
}
index++;
}
return root;
}
static int res = 0;
static HashMap<Long, Integer> hash = new HashMap<>();
public static int pathSum(TreeNode root, int targetSum) {
hash.put(0L, 1);
dfs(0L, root, targetSum);
return res;
}
public static void dfs(Long sum, TreeNode root, int targetSum) {
// 终止
if (root == null) {
return;
}
// 递归
sum += root.val;
if (hash.containsKey(sum - targetSum) && hash.get(sum - targetSum) >= 1) {
res += hash.get(sum - targetSum);
}
hash.put((Long) sum, hash.getOrDefault(sum, 0) + 1);
dfs(sum, root.left, targetSum);
dfs(sum, root.right, targetSum);
hash.put((Long) sum, hash.get(sum) - 1);
}
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
String input = sc.nextLine();
TreeNode root = build(input);
System.out.println("输入目标和");
int targetSum = sc.nextInt();
pathSum(root, targetSum);
System.out.println("结果是" + res);
}
}