LeetCode 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]method
找到第一个大于根节点的数，其后面就是右子树，前面就是左子树
TreeNode* traversal(vector<int>& pre, int l, int r) {
if (l >= r) return nullptr; // (l, r)
TreeNode *root = new TreeNode(pre[l]);
int idx = l + 1;
for (; idx < r; idx++) {
if (pre[idx] > root->val) break;
}
root->left = traversal(pre, l + 1, idx);
root->right = traversal(pre, idx, r);
return root;
}
TreeNode* bstFromPreorder(vector<int>& preorder) {
int n = preorder.size();
return traversal(preorder, 0, n);
}
