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1382.balance-a-binary-search-tree.kt
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/*
* @lc app=leetcode id=1382 lang=kotlin
*
* [1382] Balance a Binary Search Tree
*
* https://leetcode.com/problems/balance-a-binary-search-tree/description/
*
* algorithms
* Medium (73.89%)
* Likes: 52
* Dislikes: 9
* Total Accepted: 5.2K
* Total Submissions: 7K
* Testcase Example: '[1,null,2,null,3,null,4,null,null]'
*
* Given a binary search tree, return a balanced binary search tree with the
* same node values.
*
* A binary search tree is balanced if and only if the depth of the two
* subtrees of every node never differ by more than 1.
*
* If there is more than one answer, return any of them.
*
*
* Example 1:
*
*
*
*
* Input: root = [1,null,2,null,3,null,4,null,null]
* Output: [2,1,3,null,null,null,4]
* Explanation: This is not the only correct answer, [3,1,4,null,2,null,null]
* is also correct.
*
*
*
* Constraints:
*
*
* The number of nodes in the tree is between 1 and 10^4.
* The tree nodes will have distinct values between 1 and 10^5.
*
*/
// @lc code=start
/**
* Example:
* var ti = TreeNode(5)
* var v = ti.`val`
* Definition for a binary tree node.
* class TreeNode(var `val`: Int) {
* var left: TreeNode? = null
* var right: TreeNode? = null
* }
*/
class Solution {
private var arr = IntArray(10005)
private var cnt = 0
fun balanceBST(root: TreeNode?): TreeNode? {
cnt = 0
var head = root
dfs(head)
return sortedArrayToBST(cnt)
}
fun dfs(root: TreeNode?) {
if (root?.left != null) dfs(root?.left)
arr[cnt++] = root!!.`val`
if (root?.right != null) dfs(root?.right)
}
fun sortedArrayToBST(start: Int, end: Int): TreeNode? {
if (start > end) return null
val mid = start + (end - start) / 2
var root: TreeNode? = TreeNode(arr[mid])
root?.left = sortedArrayToBST(start, mid-1)
root?.right = sortedArrayToBST(mid+1, end)
return root
}
fun sortedArrayToBST(n: Int): TreeNode? {
return sortedArrayToBST(0, n-1)
}
}
// @lc code=end