Path Sum
<p>Given the <code>root</code> of a binary tree and an integer <code>targetSum</code>, return <code>true</code> if the tree has a <strong>root-to-leaf</strong> path such that adding up all the values along the path equals <code>targetSum</code>.</p>

<p>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/0100-0199/0112.Path%20Sum/images/pathsum1.jpg" style="width: 500px; height: 356px;" />
<pre>
<strong>Input:</strong> root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
<strong>Output:</strong> true
<strong>Explanation:</strong> The root-to-leaf path with the target sum is shown.
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0100-0199/0112.Path%20Sum/images/pathsum2.jpg" />
<pre>
<strong>Input:</strong> root = [1,2,3], targetSum = 5
<strong>Output:</strong> false
<strong>Explanation:</strong> There are two root-to-leaf paths in the tree:
(1 --&gt; 2): The sum is 3.
(1 --&gt; 3): The sum is 4.
There is no root-to-leaf path with sum = 5.
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> root = [], targetSum = 0
<strong>Output:</strong> false
<strong>Explanation:</strong> Since the tree is empty, there are no root-to-leaf paths.
</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; Breadth-First Search; Binary Tree
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int targetSum) {
        return dfs(root, targetSum);
    }

    private boolean dfs(TreeNode root, int s) {
        if (root == null) {
            return false;
        }
        s -= root.val;
        if (root.left == null && root.right == null && s == 0) {
            return true;
        }
        return dfs(root.left, s) || dfs(root.right, s);
    }
}