balanced Archives - Huahua's Tech Road

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)

C++

// 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);

}

};

Solution: Greedy

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

C++

// 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;

}

};

Posts tagged as “balanced”

花花酱 LeetCode 1382. Balance a Binary Search Tree

Solution: Inorder + recursion

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Solution: Greedy

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