标签：this code pre btree ber follow getheight alc none
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 depth of the two subtrees of every node never differ by more than 1.
Example 1:
Given the following tree [3,9,20,null,null,15,7]:
3
/ 9 20
/ 15 7
Return true.
Key point of solution is that we need a helper function to get max height of left child tree and right child tree.
In each level, get height is O(N) time complexity; and considering it‘s binary tree, level number is logN, so total time complexity is O(NlogN).
1 # Definition for a binary tree node. 2 # class TreeNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 8 class Solution: 9 def isBalanced(self, root: TreeNode) -> bool: 10 def getHeight(node: TreeNode) -> int: 11 if not node: 12 return 0 13 return max(getHeight(node.left), getHeight(node.right)) + 1 14 15 if not root: 16 return True 17 left_val = getHeight(root.left) 18 right_val = getHeight(root.right) 19 if abs(left_val - right_val) > 1: 20 return False 21 return self.isBalanced(root.left) and self.isBalanced(root.right)
Evaluate child tree is balanced or not while calculation heights. Time complexity is O(N).
1 # Definition for a binary tree node. 2 # class TreeNode: 3 # def __init__(self, x): 4 # self.val = x 5 # self.left = None 6 # self.right = None 7 8 class Solution: 9 def isBalanced(self, root: TreeNode) -> bool: 10 def getHeight(node: TreeNode) -> int: 11 if not node: 12 return 0 13 left = getHeight(node.left) 14 right = getHeight(node.right) 15 if left == -1 or right == -1: 16 return -1 17 if abs(left - right) > 1: 18 return -1 19 return max(left, right) + 1 20 21 if not root: 22 return True 23 return getHeight(root) != -1
标签：this code pre btree ber follow getheight alc none
原文地址：https://www.cnblogs.com/ireneyanglan/p/11371289.html
