Trim-a-Binary-Search-Tree

Dec 05, 2017

第69天。

今天的题目是Trim a Binary Search Tree:

Given a binary search tree and the lowest and highest boundaries as L and R, trim the tree so that all its elements lies in [L, R] (R >= L). You might need to change the root of the tree, so the result should return the new root of the trimmed binary search tree.

Example 1: Input:

1

/
0 2

L = 1 R = 2

Output:

1
  \
   2

Example 2: Input:

3

/
0 4
2 / 1

L = 1 R = 3

Output:

  3
 /

2 / 1

一开始没看到时二叉排序树,然后写的就有点复杂了:

TreeNode* trimBST(TreeNode* root, int L, int R) {
    if (!root) return nullptr;
    root->left = trimBST(root->left,L,R);
    root->right = trimBST(root->right,L,R);
    if (root->val >= L && root->val <= R) return root;
    else if (root->left != nullptr && root->right != nullptr) {
        auto p = root->left;
        while(p->left) p = p->left;
        p->left = root->right;
        return root->left;
    }
    return (root->left)?root->left:root->right;
}
TreeNode* trimBST(TreeNode* root, int L, int R) {
    if (root == nullptr) return nullptr;
    if (root->val < L) return trimBST(root->right,L,R);
    if (root->val > R) return trimBST(root->left,L,R);
    root->left = trimBST(root->left,L,R);
    root->right = trimBST(root->right,L,R);
    return root;
}
LeetCodeLeetCode

Count-Numbers-with-Unique-Digits

Longest-Continuous-Increasing-Subsqeuence

comments powered by Disqus