1028.recover-a-tree-from-preorder-traversal.cpp

/*
 * @lc app=leetcode id=1028 lang=cpp
 *
 * [1028] Recover a Tree From Preorder Traversal
 */

// @lc code=start
/**
 * 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 {
  int dashCount = 0;
  int getNumber(const string &pattern, int &pos) {
    int result = 0;
    while(pos < pattern.length() && isdigit(pattern[pos])) {
      result *= 10;
      result += pattern[pos] - '0';
      pos += 1;
    }
    return result;
  }
  void getDashCount(const string &pattern, int &pos) {
    dashCount = 0;
    while(pos < pattern.length() && pattern[pos] == '-') {
      dashCount += 1;
      pos += 1;
    }
  }
  TreeNode *solve(const string &pattern, int &pos, int level) {
    if(!dashCount) getDashCount(pattern, pos);
    if(dashCount != level) return nullptr;
    dashCount = 0;
    TreeNode *root = new TreeNode(getNumber(pattern, pos));
    root->left = solve(pattern, pos, level + 1);
    root->right = solve(pattern, pos, level + 1);
    return root;
  }
public:
  TreeNode* recoverFromPreorder(string traversal) {
    TreeNode *root = new TreeNode();
    int pos = 0;
    root->val = getNumber(traversal, pos);
    root->left = solve(traversal, pos, 1);
    root->right = solve(traversal, pos, 1);
    return root;
  }
};

// Accepted
// 105/105 cases passed (0 ms)
// Your runtime beats 100 % of cpp submissions
// Your memory usage beats 55.58 % of cpp submissions (21.7 MB)
// @lc code=end