LeetCode刷题系列 -- 113. 路径总和 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：

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：

输入：root = [1,2,3], targetSum = 5

输出：[]

示例 3：

输入：root = [1,2], targetSum = 0

输出：[]

113. 路径总和 II - 力扣（Leetcode）

思路：

这种路径题，常常使用回溯法。

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 {
    
    List> result = new ArrayList<>();

    public List> pathSum(TreeNode root, int targetSum) {
        if(root == null) {
            return result;
        }
        tranverse(root, targetSum, new ArrayList<>());
        return result;
    }

    void tranverse(TreeNode root, int subSum, List subList) {
        
        subList.add(root.val);
        if(root.left == null && root.right == null) {
            if(subSum - root.val == 0) {
                result.add(new ArrayList(subList));
            }
        }

        if(root.left != null) {
            tranverse(root.left, subSum - root.val, subList);
        }
        if(root.right != null) {
            tranverse(root.right, subSum - root.val, subList);
        }
        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> result;

    vector> pathSum(TreeNode* root, int targetSum) {
        if(root == nullptr) {
            return result;
        }
        vector sub_path;
        findPath(root, targetSum, sub_path);

        return result;
    }

    void findPath(TreeNode* root, int targetSum, vector& sub_path) {
         if(root ==  nullptr) {
             return;
         }
         
         sub_path.push_back(root->val);
          
          //targetSum -= root->val;
        if(root->left == nullptr && root->right == nullptr && targetSum == root->val) {
             vector tmp = sub_path;
             result.push_back(tmp);
         }

         findPath(root->left, targetSum - root->val, sub_path);
         // sub_path.erase(sub_path.end()-1);

         // sub_path.push_back(root->val);
         findPath(root->right, targetSum - root->val, sub_path);
         // sub_path.erase(sub_path.end()-1);
         sub_path.pop_back();

    }

};