0600 - 0699
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
/**
* 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;
}
};