LeetCode（算法）- 113. 路径总和 II

/**
 * 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;
 *     }
 * }
 */
// 解决方案(1)
class Solution {
    LinkedList<List<Integer>> res = new LinkedList<>();
    LinkedList<Integer> path = new LinkedList<>();
    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        recur(root, sum);
        return res;
    }
    void recur(TreeNode root, int tar) {
        if(root == null) return;
        path.add(root.val);
        tar -= root.val;
        if(tar == 0 && root.left == null && root.right == null)
            res.add(new LinkedList(path));
        recur(root.left, tar);
        recur(root.right, tar);
        path.removeLast();
    }
}
// 解决方案(2)
class Solution {
    private int target;
    private List<List<Integer>> list;
    private List<Integer> subList;
    public List<List<Integer>> pathSum(TreeNode root, int target) {
        this.target = target;
        list = new ArrayList<>();
        subList = new ArrayList<>();
        dfs(root, 0);
        return list;
    }
    private void dfs(TreeNode node, int sum) {
        if (null == node) {
            // 为了配合演出回溯的 remove, 否则会导致删除和新增不一致
            subList.add(null);
            return;
        }
        sum += node.val;
        subList.add(node.val);
        if (null == node.right && null == node.left) {
            if (sum == target) {
                list.add(new ArrayList<>(subList));
            }
            return;
        }
        dfs(node.left, sum);
        subList.remove(subList.size() - 1);
        dfs(node.right, sum);
        subList.remove(subList.size() - 1);
    }
}

/**
 * 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>> pathSum(TreeNode* root, int sum) {
        recur(root, sum);
        return res;
    }
private:
    vector<vector<int>> res;
    vector<int> path;
    void recur(TreeNode* root, int tar) {
        if(root == nullptr) return;
        path.push_back(root->val);
        tar -= root->val;
        if(tar == 0 && root->left == nullptr && root->right == nullptr)
            res.push_back(path);
        recur(root->left, tar);
        recur(root->right, tar);
        path.pop_back();
    }
};