606. Construct String from Binary Tree

最新推荐文章于 2024-12-02 12:31:36 发布

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本文介绍了一种将二叉树转换为字符串的算法，该算法遵循先序遍历的原则，并考虑了空节点的表示方式。文章提供了两种实现思路及对应的Java代码示例。

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.

 class TreeNode {
      int val;
      TreeNode left;
      TreeNode right;
      TreeNode(int x) { val = x; }
  }

思路一：

class Solution {
    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 + ")";
    }
}

class Solution {
    public String tree2str(TreeNode t) {
        if (t == null) return "";
        String left = tree2str(t.left);
        String right = tree2str(t.right);
        if (left == "" && right == "") return t.val + "";
        return t.val + "(" + left + ")" + (right == "" ? "" : "(" + right + ")");
    }
}

思路二：

class Solution {
    public String tree2str(TreeNode t) {
        StringBuilder sb = new StringBuilder();
        return helper(t, sb);
    }
    public String helper(TreeNode t, StringBuilder sb)
    {
        if (t == null) return "";
        sb.append(t.val);
        if (t.left != null || t.right != null)
        {
            sb.append("(");
            helper(t.left, sb);
            sb.append(")");
            if (t.right != null)
            {
                sb.append("(");
                helper(t.right, sb);
                sb.append(")");
            }
        }
        return sb.toString();
    }
}