LeetCode 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.
An empty tree is represented by "" instead of ()".

题意：从一个带括号的字符串，构建一颗二叉树。

解题思路： 本题仔细看字符串可以发现，每个root,left,right都是以root.val(left.val)(right.val)展示的。其中当left = null而right != null时，left展示为一个空的括号'()'。同时要考虑负数的情况，所以在取数字的时候，必须注意index所在位置。我们用一个stack存储当前构建好的TreeNode,每次遇到数字时，将数字构建成TreeNode,查看是否为stack为空，不为空，则查看stack中顶层元素的左子树是否已经有了，如果没有，那当前新构建的TreeNode就是它的左边的孩子，否则就是顶层元素的右孩子。遇到')'则从栈中pop出元素。最后stack中的元素就是root，返回root栈顶元素即可。

代码如下：

    public TreeNode str2tree(String s) {
        if (s == null || s.length() == 0) {
            return null;
        }
        Stack<TreeNode> stack = new Stack<>();
        for(int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == ')') {
                stack.pop();
            } else {
                if (c >= '0' && c <= '9' || c == '-') {
                    int start = i;
                    while(i + 1 < s.length() && s.charAt(i + 1) >= '0' && s.charAt(i + 1) <= '9') {
                        i++;
                    }
                    TreeNode node = new TreeNode(Integer.valueOf(s.substring(start, i + 1)));
                    if (!stack.iSEMpty()) {
                        TreeNode parent = stack.peek();
                        if (parent.left != null) {
                            parent.right = node;
                        } else {
                            parent.left = node;
                        }
                    }
                    stack.push(node);
                }
            }
        }
        if(stack.isEmpty()) {
            return null;
        }
        return stack.peek();