655. Print Binary Tree

Given the root of a binary tree, construct a 0-indexed m x n string matrix res that represents a formatted layout of the tree. The formatted layout matrix should be constructed using the following rules:

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

Return the constructed matrix res.

/**
 * 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 {
    vector<vector<string>> res;
    int maxDepth(TreeNode* node) {
        if(!node) return 0;
        return max(maxDepth(node->left), maxDepth(node->right)) + 1;
    }
    void dfs(TreeNode* node, int l, int r, int level = 0) {
        if(!node) return;
        int m = l + (r - l) / 2;
        dfs(node->left, l, m - 1, level + 1);
        res[level][m] = to_string(node->val);
        dfs(node->right, m + 1, r, level + 1);
    }
public:
    vector<vector<string>> printTree(TreeNode* root) {
        int h = maxDepth(root), pos = 0;
        res = vector<vector<string>>(h, vector<string>(pow(2,h) - 1));
        dfs(root, 0, pow(2,h) - 2);
        return res;
    }
};