536. Construct Binary Tree from String 从括号字符串中构建二叉树

［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 caseif (s == null || s.length() == 0) return null;//initialization: stackStack<TreeNode> stack = new Stack<TreeNode>();//for loop: new node, get substring and appendfor (int i = 0, j = i; i < s.length(); i++, j = i) {//get cchar c = s.charAt(i);//if c is )if (c == ')') stack.pop();else if ((c >= '0' && c <= '9') || (c == '-')) {//continuewhile (i + 1 < s.length() && s.charAt(i + 1) >= '0' && s.charAt(i + 1) <= '9') i++;//build new nodeTreeNode node = new TreeNode(Integer.valueOf(s.substring(j, i + 1)));if (!stack.isEmpty()) {TreeNode parent = stack.peek();//get left and appendif (parent.left != null) {parent.right = node;}else parent.left = node;}stack.push(node);}}//return the last rootreturn stack.peek() == null ? null : stack.pop();}
}

View Code

转载于:https://www.cnblogs.com/immiao0319/p/9612471.html

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536. Construct Binary Tree from String 从括号字符串中构建二叉树

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