Sword finger offer 55 - ii balanced binary tree

subject

Enter the root node of a binary tree , Judge whether the tree is a balanced binary tree . If the depth difference between the left and right subtrees of any node in a binary tree does not exceed 1, So it's a balanced binary tree .

Example 1:

Given binary tree [3,9,20,null,null,15,7]

    3
   / \
  9  20
    /  \
   15   7

return true .

Example 2:

Given binary tree [1,2,2,3,3,null,null,4,4]

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

return false .

Limit ：

0 <=  The number of nodes in a tree  <= 10000

Answer key

class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        def recur(root):
            if not root:
                return 0
            left = recur(root.left)
            if left == -1:
                return - 1
            right = recur(root.right)
            if right == -1:
                return -1
            return max(left , right) + 1 if abs(left - right) <= 1 else -1
        return recur(root) != -1