536. Construct Binary Tree from String

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.

对于例题中的”4(2(3)(1))(6(5))”，可知4是根节点，“2(3)(1)”是左子树，“6(5)”是右子树。可以想到应该用递归的方式建树。
这里怎么判断它的左子树的开始和结尾呢？
通过对”(）”进行计数，当遇到’(‘加1，遇到‘）’减去1。当为0时，说明括号是完全匹配的，此时它自己可以形成一个新树。剩下的就是加括号的右子树的字符串。

/** * 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) {        if(s.size()==0) return NULL;       int val=0;int i=0,j,n=s.size();       int sign=1;       if(s[0]=='-') {sign=-1;i++;}       while(i<n && s[i]!='(') {val=val*10+(s[i]-'0');i++;}//此时，s[i]是'('       TreeNode* root=new TreeNode(val*sign);       if(i==n) return root;//如果没有孩子       int num=1;j=i;       while(num!=0)       {           j++;          if(s[j]=='(') num++;          else if(s[j]==')') num--;       }//找到右子树结尾的括号')'        root->left=str2tree(s.substr(i+1,j-i-1));        if(j+2<n)//判断有没有右子树        root->right=str2tree(s.substr(j+2,n-3-j));       return root;    }};