Question

606. Construct String from Binary Tree

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Description

Given the root node of a binary tree, your task is to create a string representation of the tree following a specific set of formatting rules. The representation should be based on a preorder traversal of the binary tree and must adhere to the following guidelines:

• Node Representation: Each node in the tree should be represented by its integer value.

• Parentheses for Children: If a node has at least one child (either left or right), its children should be represented inside parentheses. Specifically:

  • If a node has a left child, the value of the left child should be enclosed in parentheses immediately following the node's value.
  • If a node has a right child, the value of the right child should also be enclosed in parentheses. The parentheses for the right child should follow those of the left child.

• Omitting Empty Parentheses: Any empty parentheses pairs (i.e., ()) should be omitted from the final string representation of the tree, with one specific exception: when a node has a right child but no left child. In such cases, you must include an empty pair of parentheses to indicate the absence of the left child. This ensures that the one-to-one mapping between the string representation and the original binary tree structure is maintained.

  In summary, empty parentheses pairs should be omitted when a node has only a left child or no children. However, when a node has a right child but no left child, an empty pair of parentheses must precede the representation of the right child to reflect the tree's structure accurately.

 

Example 1:

Input: root = [1,2,3,4]
Output: "1(2(4))(3)"
Explanation: Originally, it needs to be "1(2(4)())(3()())", but you need to omit all the empty parenthesis pairs. And it will be "1(2(4))(3)".

Example 2:

Input: root = [1,2,3,null,4]
Output: "1(2()(4))(3)"
Explanation: Almost the same as the first example, except the () after 2 is necessary to indicate the absence of a left child for 2 and the presence of a right child.

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -1000 <= Node.val <= 1000

Solutions

Solution 1

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def tree2str(self, root: Optional[TreeNode]) -> str:
    def dfs(root):
      if root is None:
        return ''
      if root.left is None and root.right is None:
        return str(root.val)
      if root.right is None:
        return f'{root.val}({dfs(root.left)})'
      return f'{root.val}({dfs(root.left)})({dfs(root.right)})'

    return dfs(root)

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  string tree2str(TreeNode* root) {
    if (!root) return "";
    if (!root->left && !root->right) return to_string(root->val);
    if (!root->right) return to_string(root->val) + "(" + tree2str(root->left) + ")";
    return to_string(root->val) + "(" + tree2str(root->left) + ")(" + tree2str(root->right) + ")";
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func tree2str(root *TreeNode) string {
  if root == nil {
    return ""
  }
  if root.Left == nil && root.Right == nil {
    return strconv.Itoa(root.Val)
  }
  if root.Right == nil {
    return strconv.Itoa(root.Val) + "(" + tree2str(root.Left) + ")"
  }
  return strconv.Itoa(root.Val) + "(" + tree2str(root.Left) + ")(" + tree2str(root.Right) + ")"
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function tree2str(root: TreeNode | null): string {
  if (root == null) {
    return '';
  }
  if (root.left == null && root.right == null) {
    return `${root.val}`;
  }
  return `${root.val}(${root.left ? tree2str(root.left) : ''})${
    root.right ? `(${tree2str(root.right)})` : ''
  }`;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::cell::RefCell;
use std::rc::Rc;
impl Solution {
  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut String) {
    if let Some(node) = root {
      let node = node.borrow();
      res.push_str(node.val.to_string().as_str());

      if node.left.is_none() && node.right.is_none() {
        return;
      }
      res.push('(');
      if node.left.is_some() {
        Self::dfs(&node.left, res);
      }
      res.push(')');
      if node.right.is_some() {
        res.push('(');
        Self::dfs(&node.right, res);
        res.push(')');
      }
    }
  }

  pub fn tree2str(root: Option<Rc<RefCell<TreeNode>>>) -> String {
    let mut res = String::new();
    Self::dfs(&root, &mut res);
    res
  }
}