Question
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
Related Topics 树 深度优先搜索 回溯 二叉树 👍 745 👎 0
Answer
//113,跟112相比需要找出所有的路径,dfs
void dfs(TreeNode* root, vector<vector<int>> &res, vector<int> &cur, int targetSum) {
if (!root)
return;
cur.push_back(root->val);
if (root && !root->left && !root->right && root->val == targetSum) {
res.push_back(cur);
}
dfs(root->left, res, cur, targetSum-root->val);
dfs(root->right, res, cur, targetSum-root->val);
cur.pop_back();
}
vector<vector<int>> pathSum(TreeNode* root, int targetSum) {
vector<vector<int>> res;
vector<int> cur;
dfs(root, res, cur, targetSum);
return res;
}