Question
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]
]
Solution
Traditional way, use DFS and recursion.
/**
* 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>> result = new ArrayList<List<Integer>>();
List<Integer> prevList = new ArrayList<Integer>();
dfs(root, sum, result, prevList);
return result;
} private void dfs(TreeNode root, int target, List<List<Integer>> result, List<Integer> prevList) {
if (root == null)
return;
prevList.add(root.val); if (root.left == null && root.right == null) {
if (root.val == target)
result.add(new ArrayList<Integer>(prevList));
} else {
List<Integer> tmpList2 = new ArrayList<Integer>(prevList);
if (root.left != null)
dfs(root.left, target - root.val, result, prevList);
if (root.right != null)
dfs(root.right, target - root.val, result, tmpList2);
} }
}