leetcode113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.

For 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]
]

解法

DFS，判断如果是根节点，并且符合条件，则加入列表中，向上一层要把新添加的元素remove掉。

/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> ret = new ArrayList<>();
List<Integer> list = new ArrayList<>();
if (root == null) {
return ret;
}

helper(root, sum, list, ret);
return ret;
}

public void helper(TreeNode root, int sum, List<Integer> list, List<List<Integer>> ret) {
if (root == null) {
return;
}
list.add(root.val);
if (root.left == null && root.right == null && sum - root.val == 0) {
ret.add(new ArrayList<Integer>(list));
list.remove(list.size() - 1);
return;
} else {
helper(root.left, sum - root.val, list, ret);
helper(root.right, sum - root.val,
a650
list, ret);
}
list.remove(list.size() - 1);
}
}