Given a binary search tree, return a balanced binary search tree with the same node values.

A binary search tree is balanced if and only if the depth of the two subtrees of every node never differ by more than 1.

If there is more than one answer, return any of them.

Example 1:

Input: root = [1,null,2,null,3,null,4,null,null]
Output: [2,1,3,null,null,null,4]
Explanation: This is not the only correct answer, [3,1,4,null,2,null,null] is also correct.

Constraints:

• The number of nodes in the tree is between 1 and 10^4.
• The tree nodes will have distinct values between 1 and 10^5.

Solution: Inorder + recursion

Use inorder traversal to collect a sorted array from BST. And then build a balanced BST from this sorted array in O(n) time.

Time complexity: O(n)
Space complexity: O(n)

// Author: Huahua
class Solution {
public:
  TreeNode* balanceBST(TreeNode* root) {
    vector<int> vals;
    function<void(TreeNode*)> inorder = [&](TreeNode* root) {
      if (!root) return;
      inorder(root->left);
      vals.push_back(root->val);
      inorder(root->right);
    };
    
    function<TreeNode*(int, int)> build = [&](int l, int r) {
      if (l > r) return (TreeNode*)nullptr;
      int m = l + (r - l) / 2;
      auto root = new TreeNode(vals[m]);
      root->left = build(l, m - 1);
      root->right = build(m + 1, r);
      return root;
    };
    
    inorder(root);
    return build(0, vals.size() - 1);
  }
};

The post 花花酱 LeetCode 1382. Balance a Binary Search Tree appeared first on Huahua's Tech Road.

Given a balanced string s split it in the maximum amount of balanced strings.

Return the maximum amount of splitted balanced strings.

Example 1:

Input: s = "RLRRLLRLRL"
Output: 4
Explanation: s can be split into "RL", "RRLL", "RL", "RL", each substring contains same number of 'L' and 'R'.

Example 2:

Input: s = "RLLLLRRRLR"
Output: 3
Explanation: s can be split into "RL", "LLLRRR", "LR", each substring contains same number of 'L' and 'R'.

Example 3:

Input: s = "LLLLRRRR"
Output: 1
Explanation: s can be split into "LLLLRRRR".

Constraints:

• 1 <= s.length <= 1000
• s[i] = 'L' or 'R'

Solution: Greedy

Time complexity: O(n)
Space complexity: O(1)

// Author: Huahua
class Solution {
public:
  int balancedStringSplit(string s) {
    int b = 0;
    int ans = 0;
    for (char c : s) {
      b += c == 'L' ? 1 : -1;
      if (b == 0) ++ans;
    }
    return ans;
  }
};

The post 花花酱 LeetCode 1221. Split a String in Balanced Strings appeared first on Huahua's Tech Road.