Given the root of a binary tree and an integer targetSum, return all root-to-leaf paths where the sum of the node values in the path equals targetSum. Each path should be returned as a list of the node values, not node references.
A root-to-leaf path is a path starting from the root and ending at any leaf node. A leaf is a node with no children.
Example 1:
Input: root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
Output: [[5,4,11,2],[5,8,4,5]]
Explanation: There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
Example 2:
Input: root = [1,2,3], targetSum = 5
Output: []
Example 3:
Input: root = [1,2], targetSum = 0
Output: []
Constraints:
- The number of nodes in the tree is in the range [0, 5000].
- -1000 <= Node.val <= 1000
- -1000 <= targetSum <= 1000
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> result = new LinkedList<List<Integer>>();
List<Integer> currentResult = new LinkedList<Integer>();
pathSum(root,sum,currentResult,result);
return result;
}
private void pathSum(TreeNode root, int sum, List<Integer> currentResult, List<List<Integer>> result) {
if(root==null){
return;
}
currentResult.add(new Integer(root.val)); // 这里先放入，记得移除
if(root.left == null && root.right == null && sum == root.val){
result.add(new LinkedList(currentResult));
currentResult.remove(currentResult.size() - 1);//don't forget to remove the last integer
return;
}else{
pathSum(root.left, sum - root.val, currentResult, result);
pathSum(root.right, sum - root.val, currentResult, result);
}
currentResult.remove(currentResult.size() - 1);
}
}