leetcood学习笔记-110-平衡二叉树

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题目描述：

 

方法一：

class Solution(object):
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """
        if not root:
            return True
        if abs(self.countfloor(root.left)-self.countfloor(root.right))>1:
            return False
        else:
            return self.isBalanced(root.left) and self.isBalanced(root.right)
        
    def countfloor(self,root):
        if not root:
            return 0
        else:
            return max(self.countfloor(root.left),self.countfloor(root.right))+1

法二;效率高

class Solution(object):
    
    is_stop = False     # 为了加快计算速度, 标志是否已经有结果了
    
    def isBalanced(self, root):
        """
        :type root: TreeNode
        :rtype: bool
        """

        def get_depth_and_balance(root):
            if self.is_stop:        # 已经是非平衡二叉树, 就不需要再计算
                return 0, False

            if not root:
                return 0, True
            if not root.left and not root.right:
                return 1, True

            ldep, l_is_balanced = get_depth_and_balance(root.left)
            rdep, r_is_balanced = get_depth_and_balance(root.right)
            
            depth = max(ldep, rdep) + 1
            is_balanced = abs(ldep- rdep) <= 1 and l_is_balanced and r_is_balanced
            
            if not is_balanced:
                self.is_stop = True
            return depth, is_balanced
        
        depth, is_balanced = get_depth_and_balance(root)
        return is_balanced

 另：O(N) O(N)

class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        def height(root: TreeNode) -> int:
            if not root:
                return 0
            leftHeight = height(root.left)
            rightHeight = height(root.right)
            if leftHeight == -1 or rightHeight == -1 or abs(leftHeight - rightHeight) > 1:
                return -1
            else:
                return max(leftHeight, rightHeight) + 1

        return height(root) >= 0