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。