LeetCode.606.Construct String from Binary Tree 二叉树转括号字符串

题目描述

606 Construct String from Binary Tree
https://leetcode-cn.com/problems/construct-string-from-binary-tree/

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.

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解题过程

先序遍历序列加括号，括住每个结点的左右子树并忽略无用空括号，递归求解
需要注意的是左子树无论是否为空都不能忽略，右子树为空时可忽略
时间复杂度 O(n)，空间复杂度 O(n)

public String tree2str(TreeNode t) {
    if (null == t) {
        return "";
    }
    if (null == t.left && null == t.right) {
        return String.valueOf(t.val);
    }
    StringBuilder sb = new StringBuilder(String.valueOf(t.val));
    sb.append("(").append(tree2str(t.left)).append(")");
    if (null != t.right) {
        sb.append("(").append(tree2str(t.right)).append(")");
    }
    return sb.toString();
}

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GitHub代码

algorithms/leetcode/leetcode/_606_ConstructStringFromBinaryTree.java
https://github.com/masikkk/algorithms/blob/master/leetcode/leetcode/_606_ConstructStringFromBinaryTree.java

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