LeetCode 1008 Construct Binary Search Tree from Preorder Traversal【二叉树，BST】

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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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;
    }
};