Balanced Binary Tree
Created: February 15, 2020 by [lek-tin]
Last updated: February 15, 2020
Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as:
a binary tree in which the left and right subtrees of every node differ in height by no more than
1.
Example 1
Given the following tree [3,9,20,null,null,15,7]:
3
/ \
9 20
/ \
15 7
Return true.
Example 2
Given the following tree [1,2,2,3,3,null,null,4,4]:
1
/ \
2 2
/ \
3 3
/ \
4 4
Return false.
Solution
Top-down recursion
Time complexity: O(nlogn). We can calculate this using T(h) = T(h−1) + T(h−2) + 1, height denoted as h.
Space complexity: O(n)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
if not root:
return True
lh = self.getHeight(root.left)
rh = self.getHeight(root.right)
if abs(lh-rh) <= 1 and self.isBalanced(root.left) and self.isBalanced(root.right):
return True
return False
def getHeight(self, root):
if not root:
return 0
return max(self.getHeight(root.left), self.getHeight(root.right)) + 1
Bottom-up recurssion
Time complexity: O(n).
Space complexity: O(n)
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isBalanced(self, root: TreeNode) -> bool:
return self.traverse(root)[0]
def traverse(self, root):
if not root:
return True, -1
isLeftTreeBalanced, leftHeight = self.traverse(root.left)
if not isLeftTreeBalanced:
return False, 0
isRightTreeBalanced, rightHeight = self.traverse(root.right)
if not isRightTreeBalanced:
return False, 0
return abs(leftHeight-rightHeight)<=1, max(leftHeight, rightHeight)+1