Path Sum II

Source

• leetcode: 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]
]

Java

    public List<List<Integer>> pathSum(TreeNode root, int sum) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        dfs(root, sum, new ArrayList<Integer>(), result);
        return result;
    }

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

Python

# 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]]
        """
        result = []
        self.dfs(root, sum, [], result)
        return result;

    def dfs(self, root, sum, path, result):
        if root is None: 
            return;

        if root.left is None and root.right is None:
            if root.val==sum:
                p = path[:]
                p.append(sum)
                result.append(p)
            return

        path.append(root.val)
        self.dfs(root.left, sum-root.val, path, result)
        self.dfs(root.right, sum-root.val, path, result)
        path.pop()