标题:113路径总和II-中等
题目
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
代码Java
public List<List<Integer>> pathSum(TreeNode root, int targetSum) {
List<List<Integer>> ans = new ArrayList<>();
if (root == null) return ans;
List<Integer> temp = new LinkedList<>();
int target = targetSum;
find(root, ans, temp, target);
return ans;
}
public void find(TreeNode root, List<List<Integer>> ans, List<Integer> temp, int target) {
temp.add(root.val);
if (root.left == null && root.right == null) {
int sum = 0;
for (Integer x : temp) {
sum += x;
}
if (sum == target) {
ans.add(new ArrayList<>(temp));
}
return;
}
if (root.left != null) {
find(root.left, ans, temp, target);
temp.remove(temp.size() - 1);
}
if (root.right != null) {
find(root.right, ans, temp, target);
temp.remove(temp.size() - 1);
}
}
public List<List<Integer>> pathSum1(TreeNode root, int targetSum) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
List<Integer> path = new LinkedList<>();
preorderdfs(root, targetSum, res, path);
return res;
}
public void preorderdfs(TreeNode root, int targetsum, List<List<Integer>> res, List<Integer> path) {
path.add(root.val);
if (root.left == null && root.right == null) {
if (targetsum - root.val == 0) {
res.add(new ArrayList<>(path));
}
return;
}
if (root.left != null) {
preorderdfs(root.left, targetsum - root.val, res, path);
path.remove(path.size() - 1);
}
if (root.right != null) {
preorderdfs(root.right, targetsum - root.val, res, path);
path.remove(path.size() - 1);
}
}
|