1008. Construct Binary Search Tree from Preorder Traversal

Return the root node of a binary search tree that matches the given preorder traversal.

(Recall that a binary search tree is a binary tree where for every node, any descendant of node.left has a value < node.val, and any descendant of node.right has a value > node.val.  Also recall that a preorder traversal displays the value of the node first, then traverses node.left, then traverses node.right.)

Example 1:

Input: [8,5,1,7,10,12]
Output: [8,5,10,1,7,null,12]

Note: 

1. 1 <= preorder.length <= 100
2. The values of preorder are distinct.

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根据二叉搜索树的先序遍历结果，构造该二叉搜索树。

简单题。先序遍历是根、左、右，又因为是二叉搜索树，所以左<根<右，所以数组中第一个元素是根节点，然后找到数组右边第一个大于根节点的树，把数组划分为左子树和右子树，然后递归构造。

完整代码如下：

class Solution {
public:
	TreeNode* Work(vector<int> &preorder, int l, int r) {
		if (l > r)return NULL;
		if (l == r)return new TreeNode(preorder[l]);
		TreeNode *root = new TreeNode(preorder[l]);
		int mid = -1;
		for (int i = l; i <= r; ++i) {
			if (preorder[i] > root->val) {
				mid = i;
				break;
			}
		}
		if (mid == -1)root->left = Work(preorder, l + 1, r);
		else {
			root->left = Work(preorder, l + 1, mid - 1);
			root->right = Work(preorder, mid, r);
		}
		return root;
	}
	TreeNode* bstFromPreorder(vector<int>& preorder) {
		return Work(preorder, 0, preorder.size() - 1);
	}
};

本代码提交AC，用时8MS。

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。