108. Convert Sorted Array to Binary Search Tree
Problem
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
Example
Given the sorted array: [-10,-3,0,5,9],
One possible answer is: [0,-3,9,-10,null,5], which represents the following height balanced BST:
0
/ \
-3 9
/ /
-10 5
[1] Code (20. 11. 18)
public TreeNode sortedArrayToBST(int[] nums) {
return makeBST(nums, 0, nums.length - 1);
}
private TreeNode makeBST(int[] nums, int left, int right) {
if (left > right) {
return null;
}
int centerIdx = left + (right - left) / 2; // [1]
// int centerIdx = (left + right) / 2;
// = int centerIdx = (left + right) >> 1;
TreeNode treeNode = new TreeNode(nums[centerIdx]);
treeNode.left = makeBST(nums, left, centerIdx - 1);
treeNode.right = makeBST(nums, centerIdx + 1, right);
return treeNode;
}
-
[1] : (left + right) / 2 로 할 수 있지만
이러면 overflow가 발생할 수 있다.
그러므로 left + (right - left) / 2 로 하는게 조금 더 안전하다.
[2] Code (21. 08. 29)
class Solution {
public TreeNode sortedArrayToBST(int[] nums) {
return solve(nums, 0, nums.length - 1);
}
public TreeNode solve(int[] nums, int stIdx, int endIdx) {
int midIdx = (stIdx + endIdx) / 2;
TreeNode node = new TreeNode(nums[midIdx]);
if (stIdx <= midIdx - 1) {
node.left = solve(nums, stIdx, midIdx - 1);
}
if (midIdx + 1 <= endIdx) {
node.right = solve(nums, midIdx + 1, endIdx);
}
return node;
}
}
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이전과 같은 방식으로 접근했다.
다시 풀 필요는 없어 보인다.