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113. 路径总和 II

力扣链接(中等):https://leetcode.cn/problems/path-sum-ii

给你二叉树的根节点 root 和一个整数目标和 targetSum ,找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。

叶子节点 是指没有子节点的节点。

示例 1:

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

示例 2:

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输入:root = [1,2,3], targetSum = 5
输出:[]

示例 3:

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输入:root = [1,2], targetSum = 0
输出:[]

提示:

  • 树中节点总数在范围 [0, 5000]
  • -1000 <= Node.val <= 1000
  • -1000 <= targetSum <= 1000

个人题解

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:
    vector<vector<int>> result;
    vector<int> path;
    void traversal(TreeNode* node, int count) {
        // 处理叶子节点
        if(!node->left && !node->right && count == 0) {
            result.push_back(path);
            return;
        }
        if(!node->left && !node->right) return;

        // 递归
        if(node->left) {
            path.push_back(node->left->val);
            count -= node->left->val;
            traversal(node->left, count);
            count += node->left->val;
            path.pop_back();
        }
        if(node->right) {
            path.push_back(node->right->val);
            count -= node->right->val;
            traversal(node->right, count);
            count += node->right->val;
            path.pop_back();
        }
    }

    vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
        if(!root) return vector<vector<int>>();
        path.push_back(root->val);
        traversal(root, targetSum - root->val);
        return result;
    }
};