606. Construct String from Binary Tree

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606. Construct String from Binary Tree

题目描述

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.

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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)".
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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.

题目大意

将二叉树转换为字符串，转换规则为root(left)(right)，以此类推

解题思路

递归
ps:不知道该咋说，代码一目了然。

代码

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public var val: Int
 *     public var left: TreeNode?
 *     public var right: TreeNode?
 *     public init(_ val: Int) {
 *         self.val = val
 *         self.left = nil
 *         self.right = nil
 *     }
 * }
 */
class Solution {
    func tree2str(_ t: TreeNode?) -> String {
        guard let tree = t else { return "" }
        if tree.left == nil && tree.right == nil { return "\(tree.val)" }
        else if tree.left == nil { return "\(tree.val)" + "()" + "(\(tree2str(tree.right)))" }
        else if tree.right == nil { return "\(tree.val)" + "(\(tree2str(tree.left)))" }
        else { return "\(tree.val)" + "(\(tree2str(tree.left)))" + "(\(tree2str(tree.right)))" }
    }
}
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