Print Binary Tree
<p>Given the <code>root</code> of a binary tree, construct a <strong>0-indexed</strong> <code>m x n</code> string matrix <code>res</code> that represents a <strong>formatted layout</strong> of the tree. The formatted layout matrix should be constructed using the following rules:</p>

<ul>
	<li>The <strong>height</strong> of the tree is <code>height</code>&nbsp;and the number of rows <code>m</code> should be equal to <code>height + 1</code>.</li>
	<li>The number of columns <code>n</code> should be equal to <code>2<sup>height+1</sup> - 1</code>.</li>
	<li>Place the <strong>root node</strong> in the <strong>middle</strong> of the <strong>top row</strong> (more formally, at location <code>res[0][(n-1)/2]</code>).</li>
	<li>For each node that has been placed in the matrix at position <code>res[r][c]</code>, place its <strong>left child</strong> at <code>res[r+1][c-2<sup>height-r-1</sup>]</code> and its <strong>right child</strong> at <code>res[r+1][c+2<sup>height-r-1</sup>]</code>.</li>
	<li>Continue this process until all the nodes in the tree have been placed.</li>
	<li>Any empty cells should contain the empty string <code>&quot;&quot;</code>.</li>
</ul>

<p>Return <em>the constructed matrix </em><code>res</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0655.Print%20Binary%20Tree/images/print1-tree.jpg" style="width: 141px; height: 181px;" />
<pre>
<strong>Input:</strong> root = [1,2]
<strong>Output:</strong> 
[[&quot;&quot;,&quot;1&quot;,&quot;&quot;],
&nbsp;[&quot;2&quot;,&quot;&quot;,&quot;&quot;]]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://fastly.jsdelivr.net/gh/doocs/leetcode@main/solution/0600-0699/0655.Print%20Binary%20Tree/images/print2-tree.jpg" style="width: 207px; height: 302px;" />
<pre>
<strong>Input:</strong> root = [1,2,3,null,4]
<strong>Output:</strong> 
[[&quot;&quot;,&quot;&quot;,&quot;&quot;,&quot;1&quot;,&quot;&quot;,&quot;&quot;,&quot;&quot;],
&nbsp;[&quot;&quot;,&quot;2&quot;,&quot;&quot;,&quot;&quot;,&quot;&quot;,&quot;3&quot;,&quot;&quot;],
&nbsp;[&quot;&quot;,&quot;&quot;,&quot;4&quot;,&quot;&quot;,&quot;&quot;,&quot;&quot;,&quot;&quot;]]
</pre>

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

<ul>
	<li>The number of nodes in the tree is in the range <code>[1, 2<sup>10</sup>]</code>.</li>
	<li><code>-99 &lt;= Node.val &lt;= 99</code></li>
	<li>The depth of the tree will be in the range <code>[1, 10]</code>.</li>
</ul>
Tree; Depth-First Search; Breadth-First Search; Binary Tree
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<vector<string>> printTree(TreeNode* root) {
        int h = height(root);
        int m = h + 1, n = (1 << (h + 1)) - 1;
        vector<vector<string>> ans(m, vector<string>(n, ""));
        dfs(root, ans, h, 0, (n - 1) / 2);
        return ans;
    }

    void dfs(TreeNode* root, vector<vector<string>>& ans, int h, int r, int c) {
        if (!root) return;
        ans[r][c] = to_string(root->val);
        dfs(root->left, ans, h, r + 1, c - pow(2, h - r - 1));
        dfs(root->right, ans, h, r + 1, c + pow(2, h - r - 1));
    }

    int height(TreeNode* root) {
        if (!root) return -1;
        return 1 + max(height(root->left), height(root->right));
    }
};