Path Sum 2

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

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def pathSum(self, root, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        if root is None:
            return []
        result = []
        self.dfs(root, sum, result, [])
        return result

    def dfs(self, root, sum, result, tmp):
        if root is None:
            return
        if root.left is None and root.right is None and root.val == sum:
            tmp.append(root.val)
            result.append(list(tmp))
            tmp.pop()
        else:
            tmp.append(root.val)
            if root.left:
                self.dfs(root.left, sum - root.val, result, tmp)
            if root.right:
                self.dfs(root.right, sum - root.val, result, tmp)
            tmp.pop()

/**
 * 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<>();
        if(root == null)
            return result;
        dfs(root, sum, new ArrayList<Integer>(), result);
        return result;
    }

    public void dfs(TreeNode root, int sum, List<Integer> path, List<List<Integer>> result) {
        path.add(root.val);
        if(root.left == null && root.right == null && root.val == sum) {
            result.add(new ArrayList<>(path));
        } else {
            if(root.left != null )
                dfs(root.left, sum - root.val, path, result);
            if(root.right != null)
                dfs(root.right, sum - root.val, path, result);
        }
        path.remove(path.size() - 1);
    }
}