<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://assets.leetcode.com/uploads/2021/05/03/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://assets.leetcode.com/uploads/2021/05/03/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>

/**
 * 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:
    int height(TreeNode* root){
    if(root==nullptr)return 0;
    int l=height(root->left);
    int r=height(root->right);
    return max(l,r)+1;
}
int power(int b,int a){
    if(a<0)return 1;
    if(a==1)return b;
    if(a==0)return 1;
    if(a%2==0){
        return power(b*b,a/2);
    }
    return b*power(b*b,(a-1)/2);
}

void preorder(int r,int c,TreeNode* root,vector<vector<string>>&ans,int h){
    if(root==nullptr)return;
    ans[r][c]=to_string(root->val);
    preorder(r+1,c-power(2,h-r-1),root->left,ans,h);
    preorder(r+1,c+power(2,h-r-1),root->right,ans,h);
}
vector<vector<string>> printTree(TreeNode* root) {
    int h=height(root)-1;
    int col;
    col=power(2,h+1)-1;
    vector<vector<string>>ans(h+1,vector<string>(col,""));
    preorder(0,(col-1)/2,root,ans,h);
    return ans;
}
};
10800