LeetCode刷题 Day23
669. Trim a Binary Search Tree
Given the root of a binary search tree and the lowest and highest boundaries as low and high, trim the tree so that all its elements lies in [low, high]. Trimming the tree should not change the relative structure of the elements that will remain in the tree (i.e., any node's descendant should remain a descendant). It can be proven that there is a unique answer.
Return the root of the trimmed binary search tree. Note that the root may change depending on the given bounds.
Example 1:
Input: root = [1,0,2], low = 1, high = 2
Output: [1,null,2]
Example 2:
Input: root = [3,0,4,null,2,null,null,1], low = 1, high = 3
Output: [3,2,null,1]
思路:
- 这个题目的重点在于当找到超过范围的root.val的时候如何处理，在遇到超过范围的值时应该往相反的方向继续去递归，因为我们无法判断子树的情况，是需要继续递归去判定
代码:
var trimBST = function(root, low, high) {
if (!root) return null;
if (root.val < low) return trimBST(root.right, low, high);
if (root.val > high) return trimBST(root.left, low, high);
root.left = trimBST(root.left, low, high);
root.right = trimBST(root.right, low, high);
return root;
};
时间复杂度: 最差情况 O(N) 空间复杂度: 最差情况 O(N)
108. Convert Sorted Array to Binary Search Tree
Given an integer array nums where the elements are sorted in ascending order, convert it to a height-balanced binary search tree.
Example 1:
Input: nums = [-10,-3,0,5,9]
Output: [0,-3,9,-10,null,5]
Explanation: [0,-10,5,null,-3,null,9] is also accepted:
Example 2:
Input: nums = [1,3]
Output: [3,1]
Explanation: [1,null,3] and [3,1] are both height-balanced BSTs.
思路:
- 这道题就是很典型的构造BST，通过不断的分割左右区间来获得左右子树构造BST
代码:
var sortedArrayToBST = function(nums) {
let left = 0;
let right = nums.length - 1;
var helper = function(nums, left, right) {
if (left > right) return null;
let mid = left + Math.floor((right - left) / 2);
const treeNode = new TreeNode(nums[mid]);
treeNode.left = helper(nums, left, mid - 1);
treeNode.right = helper(nums, mid + 1, right);
return treeNode;
}
return helper(nums, left, right);
};
时间复杂度: O(N) 空间复杂度: O(logN) 这个肯定是一个平衡树
538. Convert BST to Greater Tree
Given the root of a Binary Search Tree (BST), convert it to a Greater Tree such that every key of the original BST is changed to the original key plus the sum of all keys greater than the original key in BST.
As a reminder, a binary search tree is a tree that satisfies these constraints:
- The left subtree of a node contains only nodes with keys less than the node's key.
- The right subtree of a node contains only nodes with keys greater than the node's key.
- Both the left and right subtrees must also be binary search trees.
Example 1:
Input: root = [4,1,6,0,2,5,7,null,null,null,3,null,null,null,8]
Output: [30,36,21,36,35,26,15,null,null,null,33,null,null,null,8]
Example 2:
Input: root = [0,null,1]
Output: [1,null,1]
思路:
- 因为是加上比自己小的数，所以就用BST反向顺序的遍历，同时记录之前的sum
代码:
var convertBST = function(root) {
let prev = 0;
var helper = function(root) {
if (!root) return 0;
helper(root.right);
root.val = root.val + prev;
prev = root.val;
helper(root.left);
}
helper(root);
return root;
};
时间复杂度: O(N) 空间复杂度: O(N)
