LeetCode 536----Construct Binary Tree from String

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:

1. There will only be '(', ')', '-' and '0' ~ '9' in the input string.

算法分析：

在 class Solution 内声明一个 int index=0，用来标记当前应该考察的输入字符串的下标，如果下标所指示的是数字，则读取数字，并增进下标；如果下标所指示的是'('，则需要递归调用生成TreeNode的函数，并将结果放在root.left子树中；如果当前字符为')',则当前递归调用结束，将 index 增加 1 ，并返回生成的 TreeNode。

在父级的函数调用中，将返回的TreeNode放在root.left，考差当前 index 所指示的字符，如果为'('，则继续递归调用函数来生成右子树；如果为')',说明当前子树没有右子树，将 index 增加 1，并返回当前TreeNode。

Java 算法实现：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    private int index=0;
    public TreeNode str2tree(String s) {
        int len=s.length();
        if(len<1||index>=len){
        	return null;
        }
        int val=0;
        int ch;
        int sign=1;
        if(s.charAt(index)=='-'){
        	sign=-1;
        	index++;
        }
        while(index<len&&(ch=s.charAt(index))<='9'&&ch>='0'){
            val*=10;
        	val+=ch-'0';
        	index++;
        }
        TreeNode root=new TreeNode(sign*val);
        if(index>=len||s.charAt(index)==')'){
        	index++;
        	return root;//have no child
        }//here now index is pointing to a '('

        index++;//now pointing to a number
        root.left=str2tree(s);
        if(index>=len||s.charAt(index)==')'){
        	index++;
        	return root;
        }//here it means index is pointing to '('
        index++;
        root.right=str2tree(s);
        if(index>=len||s.charAt(index)==')'){
        	index++;
        }
        return root;
    }
}