1382. Balance a Binary Search Tree

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.

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给定一棵二叉搜索树，要求把这棵二叉搜索树进行平衡化处理。一棵平衡二叉搜索树首先要是一棵二叉搜索树，其次要是平衡的，平衡的意思是左右子树的高度差不超过1。

要构造一棵平衡二叉搜索树，其树的根节点应该是整个有序数组的中位数。所以先对原来的二叉搜索树先序遍历，得到有序数组，然后取数组中位数作为根节点，在左右子数组中递归构造二叉搜索树。

完整代码如下：

class Solution {
public:
	void preIter(TreeNode* root, vector<int> &nums) {
		if (root == NULL)return;
		preIter(root->left, nums);
		nums.push_back(root->val);
		preIter(root->right, nums);
	}
	TreeNode* constructBST(vector<int> &nums, int l, int r) {
		if (l > r)return NULL;
		if (l == r)return new TreeNode(nums[l]);

		int mid = l + (r - l) / 2;
		TreeNode* root = new TreeNode(nums[mid]);
		root->left = constructBST(nums, l, mid - 1);
		root->right = constructBST(nums, mid + 1, r);
		return root;
	}
	TreeNode* balanceBST(TreeNode* root) {
		vector<int> nums;
		preIter(root, nums);
		int n = nums.size();
		return constructBST(nums, 0, n - 1);
	}
};

本代码提交AC，用时180MS。