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 <= preorder.length <= 100
- The values of
preorder
are distinct.
题意
已知 BST
前序遍历序列,求对应的根节点
思路1
- BST 中序有序
- 根据中序和前序构造二叉树
代码1
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode* bstFromPreorder(vector<int>& preorder) {
vector<int> inorder(preorder);
sort(inorder.begin(), inorder.end());
TreeNode *root = build(preorder, inorder, 0, preorder.size()-1, 0, preorder.size()-1);
return root;
}
TreeNode* build(vector<int> &preorder, vector<int> &inorder, int pl, int pr, int il, int ir){
if(pl > pr) return NULL;
TreeNode *node = new TreeNode(preorder[pl]);
int k = 0;
for(int i = il; i <= ir; i++)
if(inorder[i] == node->val)
{
k = i;
break;
}
node->left = build(preorder, inorder, pl + 1, pl + 1 + k - il - 1, il, k - 1);
node->right = build(preorder, inorder, pl + k - il + 1, pr, k + 1, ir);
return node;
}
};