• LeetCode 536----Construct Binary Tree from String


    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:

    1. There will only be '(', ')', '-' and '0' ~ '9' in the input string.

    算法分析:

    在 class Solution 内声明一个 int index=0,用来标记当前应该考察的输入字符串的下标,如果下标所指示的是数字,则读取数字,并增进下标;如果下标所指示的是'(',则需要递归调用生成TreeNode的函数,并将结果放在root.left子树中;如果当前字符为')',则当前递归调用结束,将 index 增加 1 ,并返回生成的 TreeNode。
    在父级的函数调用中,将返回的TreeNode放在root.left,考差当前 index 所指示的字符,如果为'(',则继续递归调用函数来生成右子树;如果为')',说明当前子树没有右子树,将 index 增加 1,并返回当前TreeNode。

    Java 算法实现:

    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    public class Solution {
        private int index=0;
        public TreeNode str2tree(String s) {
            int len=s.length();
            if(len<1||index>=len){
            	return null;
            }
            int val=0;
            int ch;
            int sign=1;
            if(s.charAt(index)=='-'){
            	sign=-1;
            	index++;
            }
            while(index<len&&(ch=s.charAt(index))<='9'&&ch>='0'){
                val*=10;
            	val+=ch-'0';
            	index++;
            }
            TreeNode root=new TreeNode(sign*val);
            if(index>=len||s.charAt(index)==')'){
            	index++;
            	return root;//have no child
            }//here now index is pointing to a '('
            
            index++;//now pointing to a number
            root.left=str2tree(s);
            if(index>=len||s.charAt(index)==')'){
            	index++;
            	return root;
            }//here it means index is pointing to '('
            index++;
            root.right=str2tree(s);
            if(index>=len||s.charAt(index)==')'){
            	index++;
            }
            return root;
        }
    }
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  • 原文地址:https://www.cnblogs.com/dongling/p/6539707.html
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