0606 - Construct String from Binary Tree (Easy)
Problem Link
https://leetcode.com/problems/construct-string-from-binary-tree/
Problem Statement
Given the root of a binary tree, construct a string consisting of parenthesis and integers from a binary tree with the preorder traversal way, and return it.
Omit all the empty parenthesis pairs that do not affect the one-to-one mapping relationship between the string and the original binary tree.
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 unnecessary 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 we cannot omit the first parenthesis pair to break the one-to-one mapping relationship between the input and the output.
Constraints:
- The number of nodes in the tree is in the range
[1, 10^4]. -1000 <= Node.val <= 1000
Approach 1: DFS
- C++
- Python
- C#
- Elixir
- Java
/**
* 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) {}
* };
*/
// Time Complexity: O(N) where N is the number of the nodes in the tree
// Space Complexity: O(H) where H is the height of the tree.
// In worse case, H can be N when it is a left skewed binary tree / right skewed binary tree
class Solution {
public:
// case 1: root is nullptr -> ""
// case 2: root doesn't have left sub tree and right sub tree -> root->val
// case 3: root->left is not nullptr -> root->val + (dfs result from left sub tree)
// case 4: root->left is nullptr but root->right is not nullptr -> root->val + ()
// case 5: root->right is not nullptr -> root->val + (dfs result from right sub tree)
string tree2str(TreeNode* root) {
// handle case 1
if (!root) return "";
// we convert root->val to string here, then append results from different cases
string s = to_string(root->val);
// handle case 2
// this line is obviously not necessary
if (!root->left && !root->right) s += "";
// handle case 3
if (root->left) s += "(" + tree2str(root->left) + ")";
// handle case 4
// alternatively, you can use `else if (root->right) s += "()";`
if (!root->left && root->right) s += "()";
// handle case 5
if (root->right) s += "(" + tree2str(root->right) + ")";
return s;
}
};
# 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
# Time Complexity: O(N) where N is the number of the nodes in the tree
# Space Complexity: O(H) where H is the height of the tree.
# In worse case, H can be N when it is a left skewed binary tree / right skewed binary tree
class Solution:
# case 1: root is nullptr -> ""
# case 2: root doesn't have left sub tree and right sub tree -> root.val
# case 3: root.left is not nullptr -> root.val + (dfs result from left sub tree)
# case 4: root.left is nullptr but root.right is not nullptr -> root.val + ()
# case 5: root.right is not nullptr -> root.val + (dfs result from right sub tree)
def tree2str(self, root: Optional[TreeNode]) -> str:
# handle case 1
if root is None:
return ''
# we convert root.val to string here, then append results from different cases
s = str(root.val)
# handle case 2
# this line is obviously not necessary
if root.left is None and root.right is None:
s += ''
# handle case 3
if root.left:
s += '({})'.format(self.tree2str(root.left))
# handle case 4
# alternatively, you can use `elif root.right: s += '()'`
if root.left is None and root.right:
s += '()'
# handle case 5
if root.right:
s += '({})'.format(self.tree2str(root.right))
return s
public class Solution {
public string Tree2str(TreeNode root) {
if(root == null) return null;
string left = Tree2str(root.left);
string right = Tree2str(root.right);
string result = "";
if(left == null && right == null) {
result = $"{root.val}";
}
else if(left != null && right == null) {
result = $"{root.val}({left})";
}
else {
result = $"{root.val}({left})({right})";
}
return result;
}
}
defmodule Solution do
# Enumerate the cases:
@spec tree2str(root :: TreeNode.t | nil) :: String.t
def tree2str(%TreeNode{val: v, left: nil, right: nil}), do:
"#{v}"
def tree2str(%TreeNode{val: v, left: lch, right: nil}), do:
"#{v}(#{tree2str(lch)})"
def tree2str(%TreeNode{val: v, left: nil, right: rch}), do:
"#{v}()(#{tree2str(rch)})"
def tree2str(%TreeNode{val: v, left: lch, right: rch}), do:
"#{v}(#{tree2str(lch)})(#{tree2str(rch)})"
end
/**
* 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 {
/**Approach
* Access the elements in preorder dfs as we usually do and
* a simple add-on is the parenthesis for each child.
* Add the parenthesis to right child if:
* 1. right child is present [the case when both children exist]
* 2. left is null [the case when only right child exist]
* Note: parenthesis will not be added if child doesn't exist,
* why? -> 'cause null wouldn't be appended to our string
*/
StringBuilder sb = new StringBuilder();
public String tree2str(TreeNode root) {
f(root);
return sb.toString();
}
private void f(TreeNode root) {
if (root == null) return;
sb.append(root.val);
if (root.left == null && root.right == null) return;
sb.append("(");
f(root.left);
sb.append(")");
// Add the parenthesis to right child if
// 1. right child is present [the case when both children exist]
// 2. left is null [the case when only right child exist]
// Note: parenthesis will not be added if child doesn't exist,
// why? -> 'cause null wouldn't be appended to our string
if (root.left == null || root.right != null) {
sb.append("(");
}
f(root.right);
if (root.left == null || root.right != null) {
sb.append(")");
}
}
}