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.

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.

解法一

思路

如果忽略题目要求“删除不必要的空括号”的话,这题就是一个经典的二叉树递归解法了,即f(node) = node.value(f(node.left)(f(node.right)))

用代码来表示就是:

def tree2str(node):
    return str(node.val) + '(' + tree2str(node.left) + ')' + '(' + tree2str(node.right) + ')'

现在再来看题目要求“删除不必要的空括号”,照题目的例子,当子树不存在时,即产生空括号(),而当右子树不存在时,产生的空括号就是多余的,而左子树不存在产生的空括号则必须保留(因为当从字符串重建二叉树时,如果二叉树只有一对括号,则一定时左子树)。这个怎么处理呢?无非就是分析左右子树的存在性:

  • 本身为None:返回空字符串。
  • 左右子树都不存在:返回结点值的字符串形式。
  • 左子树不存在而右子树存在:保留左子树的空括号。
  • 左子树存在而右子树不存在:不保留右子树的空括号。
  • 左右子树均存在:正常处理。

Python

class Solution:
    def tree2str(self, t):
        """
        :type t: TreeNode
        :rtype: str
        """
        if not t:
            return ''
        if not t.left and not t.right:
            return str(t.val)
        if not t.left:
            return str(t.val) + '()' + '(' + self.tree2str(t.right) + ')'
        if not t.right:
            return str(t.val) + '(' + self.tree2str(t.left) + ')'
        return str(t.val) + '(' + self.tree2str(t.left) + ')' + '(' + self.tree2str(t.right) + ')'

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public String tree2str(TreeNode t) {
        if (t == null) {
            return "";
        }
        if (t.left == null && t.right == null) {
            return String.valueOf(t.val);
        }
        if (t.left == null) {
            return String.valueOf(t.val) + "()" + "(" + tree2str(t.right) + ")";
        }
        if (t.right == null) {
            return String.valueOf(t.val) + "(" + tree2str(t.left) + ")";
        }
        return String.valueOf(t.val) + "(" + tree2str(t.left) + ")" + "(" + tree2str(t.right) + ")";
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    string tree2str(TreeNode *t) {
        if (t == nullptr) {
            return "";
        }
        if (t->left == nullptr && t->right == nullptr) {
            return to_string(t->val);
        }
        if (t->left == nullptr) {
            return to_string(t->val) + "()" + "(" + tree2str(t->right) + ")";
        }
        if (t->right == nullptr) {
            return to_string(t->val) + "(" + tree2str(t->left) + ")";
        }
        return to_string(t->val) + "(" + tree2str(t->left) + ")" + "(" + tree2str(t->right) + ")";
    }
};