［抄题］：
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"()".
[暴力解法]：
时间分析：
空间分析：
[优化后]：
时间分析：
空间分析：
［奇葩输出条件］：
［奇葩corner case］：
注意下：取数可以多取几位，i+1位是数字时就继续i++
［思维问题］：
感觉我在背题：几天不背，功力全无。substring都忘了。
［英文数据结构或算法，为什么不用别的数据结构或算法］：
［一句话思路］：
［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：
［画图］：
［一刷］：
new TreeNode(Integer.valueOf(s.substring(j, i + 1)))字符串不能直接转node，需要转interger后再转node- 一直往后移用的是while循环
［二刷］：
- new TreeNode(Integer.valueOf(s.substring(j, i + 1))) j的初始值是i，计算之后也应该更新为新的i
［三刷］：
［四刷］：
［五刷］：
[五分钟肉眼debug的结果]：
［总结］：
从i j中截取字符串， j应该跟随i更新
［复杂度］：Time complexity: O() Space complexity: O()
［算法思想：迭代/递归/分治/贪心］：
［关键模板化代码］：
for (int i = 0, j = i; i < s.length(); i++, j = i) {
TreeNode node = new TreeNode(Integer.valueOf(s.substring(j, i + 1)));
}
［其他解法］：
［Follow Up］：
［LC给出的题目变变变］：
[代码风格] ：
[是否头一次写此类driver funcion的代码] ：
[潜台词] ：
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode str2tree(String s) {
//corner case
if (s == null || s.length() == 0) return null; //initialization: stack
Stack<TreeNode> stack = new Stack<TreeNode>(); //for loop: new node, get substring and append
for (int i = 0, j = i; i < s.length(); i++, j = i) {
//get c
char c = s.charAt(i); //if c is )
if (c == ')') stack.pop();
else if ((c >= '0' && c <= '9') || (c == '-')) {
//continue
while (i + 1 < s.length() && s.charAt(i + 1) >= '0' && s.charAt(i + 1) <= '9') i++;
//build new node
TreeNode node = new TreeNode(Integer.valueOf(s.substring(j, i + 1)));
if (!stack.isEmpty()) {
TreeNode parent = stack.peek();
//get left and append
if (parent.left != null) {
parent.right = node;
}
else parent.left = node;
}
stack.push(node);
} } //return the last root
return stack.peek() == null ? null : stack.pop();
}
}