LeetCode 536. Construct Binary Tree from String

原题链接在这里：https://leetcode.com/problems/construct-binary-tree-from-string/description/

题目：

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:

1. There will only be '(', ')', '-' and '0' ~ '9' in the input string.
2. An empty tree is represented by "" instead of "()".

题解：

找到第一个"(". 前面的就是root的值. 

下面第一个括号内的就是left child. 通过count来标记是否找到这层括号结束的位置. 遇到"(", count++, 遇到")", count--.

When count is back to 0, that means we find the string within first bracket, which is left child.

If now, current index j is still within lenght of string, then the rest part before s.length()-1 is right child

Time Complexity: O(s.length * h). h is tree height. 每个char可能被走过h遍. h是括号层的深度.

Space: O(h). stack space.

AC Java:

/**
 * 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) {
        if(s == null || s.length() == 0){
            return null;
        }
        
        int leftChildOpenBracket = s.indexOf("(");
        int rootVal = leftChildOpenBracket == -1 ? Integer.valueOf(s) : Integer.valueOf(s.substring(0, leftChildOpenBracket));
        TreeNode root = new TreeNode(rootVal);
        
        if(leftChildOpenBracket == -1){
            return root;
        }

        int leftCount = 0;
        int start = leftChildOpenBracket;
        for(int i = start; i<s.length(); i++){
            if(s.charAt(i) == '('){
                leftCount++;
            }else if(s.charAt(i) == ')'){
                leftCount--;
            }
            
            if(leftCount==0 && start==leftChildOpenBracket){
                root.left = str2tree(s.substring(start+1, i));
                start = i+1;
            }else if(leftCount == 0){
                root.right = str2tree(s.substring(start+1, i));
            }
        }  
        return root;
    }
}

Iteration Method. Use stack to perform preorder iteration. The top of stack should be current root.

When encountering digit, get the value, create a tree node, cur. If stack is not empty, then cur node could either be stack top tree node's left child or righ child. Then also push cur node into stack.

When encountering ')', then current level in this subtree should be finished. Pop the stack.

Time Complexity: O(n).

Space: O(h). Tree height.

AC Java: 

/**
 * 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) {
        if(s == null || s.length() == 0){
            return null;
        }
        
        Stack<TreeNode> stk = new Stack<TreeNode>();
        for(int i = 0; i<s.length(); i++){
            char c = s.charAt(i);
            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 cur = new TreeNode(Integer.valueOf(s.substring(start, i+1)));
                if(!stk.isEmpty()){
                    TreeNode top = stk.peek();
                    if(top.left == null){
                        top.left = cur;
                    }else{
                        top.right = cur;
                    }
                }
                
                stk.push(cur);
            }else if(c == ')'){
                stk.pop();
            }
        }
        
        return stk.peek();
    }
}

跟上Construct String from Binary Tree.