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LeetCode 536. Construct Binary Tree from String 從帶

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摘要:題意從一個帶括號的字符串,構(gòu)建一顆二叉樹。其中當而時,展示為一個空的括號。同時要考慮負數(shù)的情況,所以在取數(shù)字的時候,必須注意所在位置。遇到則從棧中出元素。最后中的元素就是,返回棧頂元素即可。

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(gòu)建一顆二叉樹。

解題思路: 本題仔細看字符串可以發(fā)現(xiàn),每個root,left,right都是以root.val(left.val)(right.val)展示的。其中當left = nullright != null時,left展示為一個空的括號"()"。同時要考慮負數(shù)的情況,所以在取數(shù)字的時候,必須注意index所在位置。我們用一個stack存儲當前構(gòu)建好的TreeNode,每次遇到數(shù)字時,將數(shù)字構(gòu)建成TreeNode,查看是否為stack為空,不為空,則查看stack中頂層元素的左子樹是否已經(jīng)有了,如果沒有,那當前新構(gòu)建的TreeNode就是它的左邊的孩子,否則就是頂層元素的右孩子。遇到")"則從棧中pop出元素。最后stack中的元素就是root,返回root棧頂元素即可。

代碼如下:

    public TreeNode str2tree(String s) {
        if (s == null || s.length() == 0) {
            return null;
        }
        Stack 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();
  

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