给定一个二叉树,确定它是高度平衡的。对于这个问题,一棵高度平衡的二叉树的定义是:一棵二叉树中每个节点的两个子树的深度相差不会超过1
时间复杂度是O(n),后序遍历
参考文章二叉树
def isBalancedTree(self, pRoot):
if pRoot is None:
return 0,True
right,isBalancedRight = self.isBalancedTree(pRoot.right)
left,isBalancedLeft = self.isBalancedTree(pRoot.left)
depth = max(left, right) +1
if isBalancedRight and isBalancedLeft:
diff = abs(left - right)
if diff <= 1:
return depth,True
return depth,False
def isBalanced(self, root):
# write your code here
depth,isba = self.isBalancedTree(root)
return isba
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