判断平衡二叉树

原创于 2018-12-25 15:20:10 发布 · 102 阅读

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"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: The root of binary tree.
    @return: True if this Binary tree is Balanced, or false.
    """
    def getmax(self,root):
        if root == None:
            return 0
        else:
            l = 1+self.getmax(root.left)
            r = 1+self.getmax(root.right)
            
        return max(l,r)
    def isBalanced(self, root):
        # write your code here
        if root == None:
            return True
      
        l =  self.getmax(root.left)
        r =  self.getmax(root.right)
        if abs(l-r) > 1:
            return False 
        
        return  self.isBalanced(root.left) and self.isBalanced(root.right)

首先要有一个求解深度的方法，先从最大树开始判断，最后递归每一个树的左子树和右子树。可能会从在小树当中，所以最后返回递归函数。