536. Construct Binary Tree from String

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最新推荐文章于 2024-11-30 03:39:34 发布

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文章标签： stack recursion tree string

本文链接：https://blog.youkuaiyun.com/wilzxu/article/details/89944329

博客围绕输入字符串构建二叉树展开，输入代表二叉树，含整数及括号对。介绍了两种构建方法，方法1使用栈，方法2采用递归，还给出了相关参考链接。

536. Construct Binary Tree from String

方法1: stack
方法2: recursion

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 “()”.

方法1: stack

思路：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* str2tree(string s) {
        if (s.empty()) return nullptr;
        stack<TreeNode*> st;
        int i = 0;
        TreeNode* head = nullptr;
        
        while (i < s.size()) {
            if (s[i] == ')') st.pop();
            else if (isdigit(s[i]) || s[i] == '-') {
                int j = i; 
                while (isdigit(s[i + 1])) i++;
                int num = stoi(s.substr(j, i - j + 1));
                TreeNode* head = new TreeNode(num);
                if (!st.empty()) {
                    if (!st.top() -> left) st.top() -> left = head;
                    else st.top() -> right = head; 
                }
                st.push(head);
            }
            i++;
        }
        return st.top();
    }
};

方法2: recursion

grandyang: https://www.cnblogs.com/grandyang/p/6793904.html
思路：