1008. Construct Binary Search Tree from Preorder Traversal

Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and return its root.

It is guaranteed that there is always possible to find a binary search tree with the given requirements for the given test cases.

A binary search tree is a binary tree where for every node, any descendant of Node.left has a value strictly less than Node.val, and any descendant of Node.right has a value strictly greater than Node.val.

A preorder traversal of a binary tree displays the value of the node first, then traverses Node.left, then traverses Node.right.

Example 1:

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

Example 2:

Input: preorder = [1,3]
Output: [1,null,3]

题目：给定一个二叉查找树的preorder排序数组，让还原出这个二叉查找树

思路，与654. Maximum Binary Tree思路很像，用栈来保存每个节点，不同的是，本题是以大小来判断左右子树，而654是以位置来判断左右子树。代码：

/**
 * 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:
    TreeNode* bstFromPreorder(vector<int>& preorder) {
        stack stk;
        TreeNode* head = new TreeNode(preorder[0]);
        stk.push(head);
        for(int i = 1; i < preorder.size(); i++){
            TreeNode* node = new TreeNode(preorder[i]);
            if(stk.top()->val > preorder[i]){
                stk.top()->left = node;
            } else {
                TreeNode* top = stk.top();
                while(!stk.empty() && stk.top()->val < preorder[i]){
                    top = stk.top();
                    stk.pop();
                }
                top->right = node;
            }
            stk.push(node);
        }
        return head;
    }
};

时间：O（N）， 空间：O（N）