You need to construct a binary tree from a string consisting of parenthesis and integers.
The whole input represents a binary tree. It contains an integer followed by zero, one or two pairs of parenthesis. The integer represents the root’s value and a pair of parenthesis contains a child binary tree with the same structure.
You always start to construct the left child node of the parent first if it exists.
Example:
Input: "4(2(3)(1))(6(5))"
Output: return the tree root node representing the following tree:
4
/ \
2 6
/ \ /
3 1 5
Note:
There will only be ‘(‘, ‘)’, ‘-‘ and ‘0’ ~ ‘9’ in the input string.
An empty tree is represented by “” instead of “()”.
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* str2tree(string s) {
int i = 0;
return dfs(s, i);
}
int get_next_num(string &s, int &i) {
int start = i;
while (i != (int)s.size() && s[i] != '(' && s[i] != ')')
i++;
return stoi(s.substr(start, i - start));
}
/*
0123456789
4(2(3)(1))(6(5))
4(2)(3)
*/
TreeNode *dfs(string &s, int &i) {
//base case
int size = s.size();
if (i >= size || s[i] == ')')
return NULL;
if (s[i] == '(')
i++;
int num = get_next_num(s, i);
TreeNode *root = new TreeNode(num);
TreeNode *left = dfs(s, i);
TreeNode *right = dfs(s, i);
root->left = left;
root->right = right;
if (s[i] == ')')
i++;
return root;
}
};