1+ //! Link:https://leetcode-cn.com/problems/balance-a-binary-search-tree
2+ use super :: * ;
3+ struct Solution { }
4+ 5+ 6+ 7+ impl Solution {
8+ pub fn balance_bst ( root : Option < Rc < RefCell < TreeNode > > > ) -> Option < Rc < RefCell < TreeNode > > > {
9+ let mut nums: Vec < i32 > = Vec :: new ( ) ;
10+ 11+ // 中序深度遍历拿到有序数组
12+ fn build_sort_vector ( r : Option < Rc < RefCell < TreeNode > > > , n : & mut Vec < i32 > ) {
13+ if let Some ( v) = r {
14+ build_sort_vector ( RefCell :: borrow ( & v) . left . clone ( ) , n) ;
15+ n. push ( RefCell :: borrow ( & v) . val ) ;
16+ build_sort_vector ( RefCell :: borrow ( & v) . right . clone ( ) , n) ;
17+ }
18+ }
19+ ;
20+ 21+ // 递归构建二叉平衡术
22+ fn build_bbst ( n : & Vec < i32 > , s : usize , e : usize ) -> Option < Rc < RefCell < TreeNode > > > {
23+ if s > e {
24+ return None
25+ }
26+ 27+ let m = ( s + e) / 2 ;
28+ let mut root = TreeNode :: new ( n[ m] ) ;
29+ if s == e {
30+ return Some ( Rc :: new ( RefCell :: new ( root) ) )
31+ }
32+ 33+ if m >= 1 {
34+ let left_node = build_bbst ( n, s, m - 1 ) ;
35+ root. left = left_node. clone ( ) ;
36+ }
37+ 38+ let right_node = build_bbst ( n, m + 1 , e) ;
39+ root. right = right_node. clone ( ) ;
40+ return Some ( Rc :: new ( RefCell :: new ( root) ) )
41+ }
42+ 43+ build_sort_vector ( root, & mut nums) ;
44+ build_bbst ( & nums, 0 , nums. len ( ) -1 )
45+ }
46+ }
47+ 48+ 49+ #[ cfg( test) ]
50+ mod tests {
51+ use super :: * ;
52+ 53+ #[ test]
54+ fn test_1 ( ) {
55+ let trees = Solution :: balance_bst ( tree ! ( 1 , null, 2 , null, 3 , null, 4 , null) ) ;
56+ println ! ( "{:?}" , trees. unwrap( ) ) ;
57+ 58+ }
59+ }
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