Sum of Root To Leaf Binary Numbers
<p>You are given the <code>root</code> of a binary tree where each node has a value <code>0</code> or <code>1</code>. Each root-to-leaf path represents a binary number starting with the most significant bit.</p>

<ul>
	<li>For example, if the path is <code>0 -&gt; 1 -&gt; 1 -&gt; 0 -&gt; 1</code>, then this could represent <code>01101</code> in binary, which is <code>13</code>.</li>
</ul>

<p>For all leaves in the tree, consider the numbers represented by the path from the root to that leaf. Return <em>the sum of these numbers</em>.</p>

<p>The test cases are generated so that the answer fits in a <strong>32-bits</strong> integer.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/1000-1099/1022.Sum%20of%20Root%20To%20Leaf%20Binary%20Numbers/images/sum-of-root-to-leaf-binary-numbers.png" style="width: 400px; height: 263px;" />
<pre>
<strong>Input:</strong> root = [1,0,1,0,1,0,1]
<strong>Output:</strong> 22
<strong>Explanation: </strong>(100) + (101) + (110) + (111) = 4 + 5 + 6 + 7 = 22
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> root = [0]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 1000]</code>.</li>
	<li><code>Node.val</code> is <code>0</code> or <code>1</code>.</li>
</ul>
Tree; Depth-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 int sumRootToLeaf(TreeNode root) {
        return dfs(root, 0);
    }

    private int dfs(TreeNode root, int t) {
        if (root == null) {
            return 0;
        }
        t = (t << 1) | root.val;
        if (root.left == null && root.right == null) {
            return t;
        }
        return dfs(root.left, t) + dfs(root.right, t);
    }
}