LeetCode 536. Construct Binary Tree from String 從帶括號字符串構建二叉樹

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 ()".

題意：從一個帶括號的字符串，構建一顆二叉樹。

解題思路： 本題仔細看字符串能夠發現，每一個root,left,right都是以root.val(left.val)(right.val)展現的。其中當left = null而right != null時，left展現爲一個空的括號'()'。同時要考慮負數的狀況，因此在取數字的時候，必須注意index所在位置。咱們用一個stack存儲當前構建好的TreeNode,每次遇到數字時，將數字構建成TreeNode,查看是否爲stack爲空，不爲空，則查看stack中頂層元素的左子樹是否已經有了，若是沒有，那當前新構建的TreeNode就是它的左邊的孩子，不然就是頂層元素的右孩子。遇到')'則從棧中pop出元素。最後stack中的元素就是root，返回root棧頂元素便可。

代碼以下：

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();