606. Construct String from Binary Tree-优快云博客

本文介绍了一种算法，用于从前序遍历的角度构建二叉树的字符串表示，并且优化了表示方式，移除了不必要的空括号对，以保持字符串与原始二叉树间的一一对应关系。

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You need to construct a string consists of parenthesis and integers from a binary tree with the preorder traversing way.

The null node needs to be represented by empty parenthesis pair "()". And you need to omit all the empty parenthesis pairs that don't affect the one-to-one mapping relationship between the string and the original binary tree.

Example 1:

Input: Binary tree: [1,2,3,4]
       1
     /   \
    2     3
   /    
  4     

Output: "1(2(4))(3)"

Explanation: Originallay it needs to be "1(2(4)())(3()())", 
but you need to omit all the unnecessary empty parenthesis pairs. 
And it will be "1(2(4))(3)".

Example 2:

Input: Binary tree: [1,2,3,null,4]
       1
     /   \
    2     3
     \  
      4 

Output: "1(2()(4))(3)"

Explanation: Almost the same as the first example, 
except we can't omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.

Seen this question in a real interview before?

Yes

题意：

按层次输入一个二叉树，输出二叉树的前序遍历顺寻，用括号包围某一个元素，去除掉不影响二叉树的空括号

算法思路：

前序遍历肯定是要用到递归的思路，根据例子上的要求，

某个节点有左子没有右子，可以省略右子的括号，如

果有右子没有左子，不可以省略左子的括号，

左右子都没有直接返回该节点。使用递归

左右子都有，返回左右子都应该带括号

代码：

package easy;

public class ConstructStringfromBinaryTree {
	public String tree2str(TreeNode t) {
        if(t == null) return "";
        
        String result = t.val + "";
        
        String left = tree2str(t.left);
        
        String right = tree2str(t.right);
        
        if(left == "" && right == ""){
        	return result;
        }
        if(left == ""){
        	return result + "()" + "(" + right + ")";
        }
        if(right == ""){
        	return result + "(" + left + ")";
        }
        
        return result + "(" + left + ")" + "(" + right + ")";
    }
	
	public class TreeNode {
		int val;
		TreeNode left;
		TreeNode right;
		
		TreeNode(int x) 
		{ val = x; }
		  }
}