[LeetCode] 536. Construct Binary Tree from String 从字符串创建二叉树
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"()"
.
这道题让我们根据一个字符串来创建一个二叉树,其中结点与其左右子树是用括号隔开,每个括号中又是数字后面的跟括号的模式,这种模型就很有递归的感觉,所以当然可以使用递归来做。首先要做的是先找出根结点值,找第一个左括号的位置,如果找不到,说明当前字符串都是数字,直接转化为整型,然后新建结点返回即可。否则的话从当前位置开始遍历,因为当前位置是一个左括号,目标是找到与之对应的右括号的位置,但是由于中间还会遇到左右括号,所以需要用一个变量 cnt 来记录左括号的个数,如果遇到左括号,cnt 自增1,如果遇到右括号,cnt 自减1,这样当某个时刻 cnt 为0的时候,就确定了一个完整的子树的位置,那么问题来了,这个子树到底是左子树还是右子树呢,需要一个辅助变量 start,当最开始找到第一个左括号的位置时,将 start 赋值为该位置,那么当 cnt 为0时,如果 start 还是原来的位置,说明这个是左子树,对其调用递归函数,注意此时更新 start 的位置,这样就能区分左右子树了,参见代码如下:
解法一:
class Solution { public: TreeNode* str2tree(string s) { if (s.empty()) return NULL; auto found = s.find('('); int val = (found == string::npos) ? stoi(s) : stoi(s.substr(0, found)); TreeNode *cur = new TreeNode(val); if (found == string::npos) return cur; int start = found, cnt = 0; for (int i = start; i < s.size(); ++i) { if (s[i] == '(') ++cnt; else if (s[i] == ')') --cnt; if (cnt == 0 && start == found) { cur->left = str2tree(s.substr(start + 1, i - start - 1)); start = i + 1; } else if (cnt == 0) { cur->right = str2tree(s.substr(start + 1, i - start - 1)); } } return cur; } };
下面这种解法使用迭代来做的,借助栈 stack 来实现。遍历字符串s,用变量j记录当前位置i,然后看当前遍历到的字符是什么,如果遇到的是左括号,什么也不做继续遍历;如果遇到的是数字或者负号,那么将连续的数字都找出来,然后转为整型并新建结点,此时我们看 stack 中是否有结点,如果有的话,当前结点就是栈顶结点的子结点,如果栈顶结点没有左子结点,那么此结点就是其左子结点,反之则为其右子结点。之后要将此结点压入栈中。如果遍历到的是右括号,说明栈顶元素的子结点已经处理完了,将其移除栈,参见代码如下:
解法二:
class Solution { public: TreeNode* str2tree(string s) { if (s.empty()) return NULL; stack<TreeNode*> st; for (int i = 0; i < s.size(); ++i) { int j = i; if (s[i] == ')') st.pop(); else if ((s[i] >= '0' && s[i] <= '9') || s[i] == '-') { while (i + 1 < s.size() && s[i + 1] >= '0' && s[i + 1] <= '9') ++i; TreeNode *cur = new TreeNode(stoi(s.substr(j, i - j + 1))); if (!st.empty()) { TreeNode *t = st.top(); if (!t->left) t->left = cur; else t->right = cur; } st.push(cur); } } return st.top(); } };
Github 同步地址:
https://github.com/grandyang/leetcode/issues/536
类似题目:
Construct String from Binary Tree
参考资料:
https://leetcode.com/problems/construct-binary-tree-from-string/
https://leetcode.com/problems/construct-binary-tree-from-string/discuss/100359/Java-stack-solution