Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
Note: A leaf is a node with no children.
Example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
Return:
[
[5,4,11,2],
[5,8,4,5]
]
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
private List<List<Integer>> res;
public List<List<Integer>> pathSum(TreeNode root, int sum) {
res = new ArrayList<>(); //动态数组
findPath(root, sum, new ArrayList<>());
return res;
}
private void findPath(TreeNode node, int sum, List<Integer> collector){
//boolean isLeaf = (node.left == null && node.right == null);//判断当前节点是否为叶子节点。
if(node == null)
return;
sum -= node.val;
collector.add(node.val);
if(sum == 0 && node.left == null && node. right == null){
res.add(new ArrayList<>(collector));
}else{
findPath(node.left, sum, collector);
findPath(node.right, sum, collector);
}
collector.remove(collector.size() - 1);
}
}/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> res;
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<int> path;//record the path.
int currentSum = 0;
FindPath(root, currentSum, path, sum);
return res;
}
void FindPath(TreeNode* root, int currentSum, vector<int>& path, int TotalSum){
if(root == NULL)
return;
currentSum += root -> val;
path.push_back(root -> val);
//decide
bool isLeaf = (root -> right == NULL && root -> left == NULL);
if(currentSum == TotalSum && isLeaf){
res.push_back(path);
path.pop_back();
return;
}
//if not, continue
if(root -> left != NULL)
FindPath(root -> left, currentSum, path, TotalSum);
if(root -> right != NULL)
FindPath(root -> right, currentSum, path, TotalSum);
//after all, pop the top stack
path.pop_back();
}
};