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.
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.
- The number of nodes in the tree is between
1
and10^4
. - The tree nodes will have distinct values between
1
and10^5
.
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
def balanceBST(self, root: TreeNode) -> TreeNode:
def foo(vals: List[int]) -> TreeNode:
if not vals:
return None
mid = len(vals) // 2
return TreeNode(vals[mid], foo(vals[:mid]), foo(vals[mid + 1:]))
curr = root
stack = []
vals = []
while stack or curr:
while curr:
stack.append(curr)
curr = curr.left
curr = stack.pop()
vals.append(curr.val)
curr = curr.right
return foo(vals)