# path-sum-ii


Try it on <a href='https://leetcode.com/problems/path-sum-ii'>leetcode</a>

## Description
<div class="description">
<div><p>Given the <code>root</code> of a binary tree and an integer <code>targetSum</code>, return <em>all <strong>root-to-leaf</strong> paths where the sum of the node values in the path equals </em><code>targetSum</code><em>. Each path should be returned as a list of the node <strong>values</strong>, not node references</em>.</p>

<p>A <strong>root-to-leaf</strong> path is a path starting from the root and ending at any leaf node. A <strong>leaf</strong> is a node with no children.</p>

<p>&nbsp;</p>
<p><strong>Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsumii1.jpg" style="width: 500px; height: 356px;">
<pre><strong>Input:</strong> root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
<strong>Output:</strong> [[5,4,11,2],[5,8,4,5]]
<strong>Explanation:</strong> There are two paths whose sum equals targetSum:
5 + 4 + 11 + 2 = 22
5 + 8 + 4 + 5 = 22
</pre>

<p><strong>Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/18/pathsum2.jpg" style="width: 212px; height: 181px;">
<pre><strong>Input:</strong> root = [1,2,3], targetSum = 5
<strong>Output:</strong> []
</pre>

<p><strong>Example 3:</strong></p>

<pre><strong>Input:</strong> root = [1,2], targetSum = 0
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the tree is in the range <code>[0, 5000]</code>.</li>
	<li><code>-1000 &lt;= Node.val &lt;= 1000</code></li>
	<li><code>-1000 &lt;= targetSum &lt;= 1000</code></li>
</ul>
</div>
</div>

## Solution(Python)
```Python
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        paths = []
        path = []

        def preorder(curr_node, sum_so_far):
            nonlocal path
            if not curr_node:
                return
            sum_so_far += curr_node.val
            path.append(curr_node.val)

            if not curr_node.left and not curr_node.right:
                if sum_so_far == targetSum:
                    paths.append(path[:])
            if curr_node.left:
                preorder(curr_node.left, sum_so_far)
            if curr_node.right:
                preorder(curr_node.right, sum_so_far)

            path.pop()

        preorder(root, 0)
        return paths

```