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606. Construct String from Binary Tree

  • Time: $O(n)$
  • Space: $O(n)$
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class Solution {
 public:
  string tree2str(TreeNode* t) {
    return dfs(t);
  }

 private:
  string dfs(TreeNode* root) {
    if (root == nullptr)
      return "";

    const string& rootStr = to_string(root->val);
    if (root->right)
      return rootStr + "(" + dfs(root->left) + ")(" + dfs(root->right) + ")";
    if (root->left)
      return rootStr + "(" + dfs(root->left) + ")";
    return rootStr + "";
  }
};

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class Solution {
  public String tree2str(TreeNode t) {
    return dfs(t);
  }

  private String dfs(TreeNode root) {
    if (root == null)
      return "";
    if (root.right != null)
      return root.val + "(" + dfs(root.left) + ")(" + dfs(root.right) + ")";
    if (root.left != null)
      return root.val + "(" + dfs(root.left) + ")";
    return root.val + "";
  }
}

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class Solution:
  def tree2str(self, t: TreeNode | None) -> str:
    def dfs(root: TreeNode | None) -> str:
      if not root:
        return ''
      if root.right:
        return str(root.val) + '(' + dfs(root.left) + ')(' + dfs(root.right) + ')'
      if root.left:
        return str(root.val) + '(' + dfs(root.left) + ')'
      return str(root.val)
    return dfs(t)