1382. Balance a Binary Search Tree

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1382. Balance a Binary Search Tree

Description

Given the root of a binary search tree, return a balanced binary search tree with the same node values. If there is more than one answer, return any of them .

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

Example 1:

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

Example 2:

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Input: root = [2,1,3]
Output: [2,1,3]

Constraints:

  • The number of nodes in the tree is in the range [1, 10^4].
  • 1 <= Node.val <= 10^5

Hints/Notes

Solution

Language: C++

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/**
 * 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:
    vector<TreeNode*> nodes;

    TreeNode* balanceBST(TreeNode* root) {
        if (!root) {
            return root;
        }
        inorderTraversal(root);

        return buildTree(0, nodes.size() - 1);
    }

    TreeNode* buildTree(int start, int end) {
        if (start > end) {
            return nullptr;
        }
        int mid = (start + end) / 2;
        TreeNode* root = nodes[mid];
        root->left = buildTree(start, mid - 1);
        root->right = buildTree(mid + 1, end);
        return root;
    }

    void inorderTraversal(TreeNode* root) {
        if (root->left) inorderTraversal(root->left);
        nodes.push_back(root);
        if (root->right) inorderTraversal(root->right);
    }
};