}vector<int>left(preorder.begin() +1, preorder.begin() + i);vector<int>right(preorder.begin() + i, preorder.end()); node->left =bstFromPreorder(left); node->right =bstFromPreorder(right);returnnode; } }; Github 同步地址: https://github.com/grandyang/leetcode/issues/1008 类似...
Given an array of integers preorder, which represents the preorder traversal of a BST (i.e., binary search tree), construct the tree and returnits 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 b...
The first element should be root. As BST, root left subtree should be smaller than root value, right subtree should be bigger than root value. Could use root value as pivot and find out array corresponding to left subtree, also array corresponding to right subtree. Time Complexity: O(nlogn...
*https://leetcode.com/problems/construct-binary-search-tree-from-preorder-traversal/description/* * 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 ...