LeetCode（剑指 Offer）- 34. 二叉树中和为某一值的路径

Posted 2022-04-26 放羊的牧码

题目链接：点击打开链接

题目大意：略

解题思路：解决方案(1) & 解决方案(2) 的区别在于 list.add & list.remove 的时机，只要能前后对称起来即可

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AC 代码

• 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;
 *     
 * 
 */
 
// 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);
    

• 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>> 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();
    
;