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

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本文介绍LeetCode上第606题Construct String from Binary Tree的两种解法，一种执行时间为31ms，另一种更快，仅需13ms。文章通过具体的例子解释了如何从二叉树的先序遍历中构造由括号和整数组成的字符串，并讨论了如何省略不影响字符串与原始二叉树一一对应关系的空括号。

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LeetCode: 606. 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.

自己的答案，执行的时间是31ms：

/**
 * 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 "";
        }
        String result = String.valueOf(t.val);
        String left = tree2str(t.left);
        String right = tree2str(t.right);
        if (!left.isEmpty() && !right.isEmpty()) {
            result += "(" + left + ")(" + right + ")";
        } else if (!left.isEmpty()) {
            result += "(" + left + ")";
        } else if (!right.isEmpty()) {
            result += "()(" + right + ")";
        }
        return result;
    }
}

速度最快的答案，执行时间是13ms：

/**
 * 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) {
        StringBuilder sb = new StringBuilder();
        dfs(t, sb);

        return sb.toString();
    }

    private void dfs(TreeNode cur, StringBuilder sb) {
        if (cur == null) {
            return;
        }

        sb.append(cur.val);

        if (cur.left != null) {
            sb.append('(');
            dfs(cur.left, sb);
            sb.append(')');
        }

        if (cur.right != null) {
            if (cur.left == null) {
                sb.append("()");
            }

            sb.append('(');
            dfs(cur.right, sb);
            sb.append(')');
        }
    }
}

Solution