README.md

剑指 Offer II 050. 向下的路径节点之和

题目描述

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）。

 

示例 1：

输入：root = [10,5,-3,3,2,null,11,3,-2,null,1], targetSum = 8
输出：3
解释：和等于 8 的路径有 3 条，如图所示。

示例 2：

输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：3

 

提示:

• 二叉树的节点个数的范围是 [0,1000]
• -109 <= Node.val <= 109 
• -1000 <= targetSum <= 1000 

 

注意：本题与主站 437 题相同：https://leetcode.cn/problems/path-sum-iii/

解法

在遍历的过程中，记录当前路径上的前缀和

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: TreeNode, targetSum: int) -> int:
        preSum = defaultdict(int)
        preSum[0] = 1

        def dfs(node: TreeNode, cur: int) -> int:
            if not node:
                return 0

            cur += node.val
            ret = preSum[cur - targetSum]

            preSum[cur] += 1
            ret += dfs(node.left, cur)
            ret += dfs(node.right, cur)
            preSum[cur] -= 1

            return ret

        return dfs(root, 0)

Java

/**
 * Definition for a binary tree node.
 * public 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;
 *     }
 * }
 */
class Solution {

    private final Map<Integer, Integer> preSum = new HashMap<>();

    public int pathSum(TreeNode root, int targetSum) {
        preSum.put(0, 1);
        return dfs(root, 0, targetSum);
    }

    private int dfs(TreeNode node, int cur, int targetSum) {
        if (node == null) {
            return 0;
        }

        cur += node.val;
        int ret = preSum.getOrDefault(cur - targetSum, 0);

        preSum.merge(cur, 1, Integer::sum);
        ret += dfs(node.left, cur, targetSum);
        ret += dfs(node.right, cur, targetSum);
        preSum.merge(cur, -1, Integer::sum);

        return ret;
    }
}

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) int {
	preSum := make(map[int]int)
	preSum[0] = 1

	var dfs func(*TreeNode, int) int
	dfs = func(node *TreeNode, cur int) int {
		if node == nil {
			return 0
		}

		cur += node.Val
		ret := preSum[cur-targetSum]

		preSum[cur]++
		ret += dfs(node.Left, cur)
		ret += dfs(node.Right, cur)
		preSum[cur]--

		return ret
	}

	return dfs(root, 0)
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int pathSum(TreeNode* root, int targetSum) {
        unordered_map<int, int> preSum;
        preSum[0] = 1;

        function<int(TreeNode*, int)> dfs = [&](TreeNode* node, int cur) {
            if (node == nullptr) {
                return 0;
            }

            cur += node->val;
            int ret = preSum[cur - targetSum];

            ++preSum[cur];
            ret += dfs(node->left, cur);
            ret += dfs(node->right, cur);
            --preSum[cur];

            return ret;
        };

        return dfs(root, 0);
    }
};

...