《剑指Offer》刷题笔记——面试题55-II. 平衡二叉树

难度：简单

一、题目描述：

二、解题分析：

1、剑指解析

I、自顶向下

II、自底向上

2、代码实现

I、自顶向下

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # Compute the tree's height via recursion
    def height(self, root: TreeNode) -> int:
        # An empty tree has height -1
        if not root:
            return -1
        return 1 + max(self.height(root.left), self.height(root.right))
    
    def isBalanced(self, root: TreeNode) -> bool:
        # An empty tree satisfies the definition of a balanced tree
        if not root:
            return True

        # Check if subtrees have height within 1. If they do, check if the
        # subtrees are balanced
        return abs(self.height(root.left) - self.height(root.right)) < 2 \
            and self.isBalanced(root.left) \
            and self.isBalanced(root.right)

II、自底向上

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
class Solution:
    # Return whether or not the tree at root is balanced while also returning
    # the tree's height
    def isBalancedHelper(self, root: TreeNode) -> (bool, int):
        # An empty tree is balanced and has height -1
        if not root:
            return True, -1
        
        # Check subtrees to see if they are balanced. 
        leftIsBalanced, leftHeight = self.isBalancedHelper(root.left)
        if not leftIsBalanced:
            return False, 0
        rightIsBalanced, rightHeight = self.isBalancedHelper(root.right)
        if not rightIsBalanced:
            return False, 0
        
        # If the subtrees are balanced, check if the current tree is balanced
        # using their height
        return (abs(leftHeight - rightHeight) < 2), 1 + max(leftHeight, rightHeight)
        
    def isBalanced(self, root: TreeNode) -> bool:
        return self.isBalancedHelper(root)[0]