Path Sum II
<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 class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0113.Path%20Sum%20II/images/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 class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0113.Path%20Sum%20II/images/pathsum2.jpg" style="width: 212px; height: 181px;" />
<pre>
<strong>Input:</strong> root = [1,2,3], targetSum = 5
<strong>Output:</strong> []
</pre>

<p><strong class="example">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>
Tree; Depth-First Search; Backtracking; Binary Tree
/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @param {number} targetSum
 * @return {number[][]}
 */
var pathSum = function (root, targetSum) {
    const ans = [];
    const t = [];
    function dfs(root, s) {
        if (!root) return;
        s -= root.val;
        t.push(root.val);
        if (!root.left && !root.right && s == 0) ans.push([...t]);
        dfs(root.left, s);
        dfs(root.right, s);
        t.pop();
    }
    dfs(root, targetSum);
    return ans;
};
