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:[]
提示:
树中节点总数在范围 [0, 5000] 内
-1000 <= Node.val <= 1000
-1000 <= targetSum <= 1000
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/path-sum-ii
方法一:DFS
C++提交内容:
class Solution {
public:
vector> ret;
vector path;
void dfs(TreeNode* root, int targetSum) {
if (root == nullptr) {
return;
}
path.emplace_back(root->val);
targetSum -= root->val;
if (root->left == nullptr && root->right == nullptr && targetSum == 0) {
ret.emplace_back(path);
}
dfs(root->left, targetSum);
dfs(root->right, targetSum);
path.pop_back();
}
vector> pathSum(TreeNode* root, int targetSum) {
dfs(root, targetSum);
return ret;
}
};



