[LeetCode] 437. Path Sum III
Given the root
of a binary tree and an integer targetSum
, return the number of paths where the sum of the values along the path equals targetSum
.
The path does not need to start or end at the root or a leaf, but it must go downwards (i.e., traveling only from parent nodes to child nodes).
Example 1:
Input: root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8 Output: 3 Explanation: The paths that sum to 8 are shown.
Example 2:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22 Output: 3
Constraints:
- The number of nodes in the tree is in the range
[0, 1000]
. -109 <= Node.val <= 109
-1000 <= targetSum <= 1000
路径总和 III。
给定一个二叉树的根节点 root ,和一个整数 targetSum ,求该二叉树里节点值之和等于 targetSum 的 路径 的数目。
路径 不需要从根节点开始,也不需要在叶子节点结束,但是路径方向必须是向下的(只能从父节点到子节点)。
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/path-sum-iii
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题意和前两个版本类似,但是这个题找的路径和,起点可以不是根节点,问满足路径和为 sum 的路径有多少。
思路是前缀和。利用前缀和的思路,递归统计你遍历到的子树上的前缀和 presum 是多少,并将这个前缀和存入 hashmap。其他的部分就是很常规的前序遍历。注意最后一个 case 会导致 Integer 越界,需要用 long 型。
时间O(n) - 树有 n 个节点
空间O(n) - hashmap
Java实现
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode() {} 8 * TreeNode(int val) { this.val = val; } 9 * TreeNode(int val, TreeNode left, TreeNode right) { 10 * this.val = val; 11 * this.left = left; 12 * this.right = right; 13 * } 14 * } 15 */ 16 class Solution { 17 int count = 0; 18 19 public int pathSum(TreeNode root, int targetSum) { 20 HashMap<Long, Integer> map = new HashMap<>(); 21 map.put(0L, 1); 22 helper(root, 0L, targetSum, map); 23 return count; 24 } 25 26 private void helper(TreeNode root, long curSum, int targetSum, HashMap<Long, Integer> map) { 27 if (root == null) { 28 return; 29 } 30 curSum += root.val; 31 if (map.containsKey(curSum - targetSum)) { 32 count += map.get(curSum - targetSum); 33 } 34 map.put(curSum, map.getOrDefault(curSum, 0) + 1); 35 helper(root.left, curSum, targetSum, map); 36 helper(root.right, curSum, targetSum, map); 37 map.put(curSum, map.get(curSum) - 1); 38 } 39 }