110. 平衡二叉树

Easy

思路

代码

python3

class Solution:
  def isBalanced(self, root: TreeNode) -> bool:
    res = []
    def dfs(node, h):
      if node is None:
        return h
      lh = dfs(node.left, h+1)
      rh = dfs(node.right, h+1)
      if abs(lh - rh) > 1:
        res.append(False)
      else:
        res.append(True)
      return max(lh,rh)
    dfs(root,0)
    return False not in res