【leetcode】110. 平衡二叉树(python)自顶向下的递归 + 自底向上的递归

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  • Post category:python


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方法一:自顶向下的递归

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isBalanced(self, root: TreeNode) -> bool:
        def dfs(node):
            if not node:
                return 0
            l = dfs(node.left)
            r = dfs(node.right)
            return max(l, r) + 1

        if not root:
            return True

        if not self.isBalanced(root.left) or not self.isBalanced(root.right):
            return False

        return True if abs(dfs(root.left) - dfs(root.right)) <= 1 else False



方法二:自底向上的递归

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def dfs(self, node):
            if not node:
                return 0
            l = self.dfs(node.left)
            r = self.dfs(node.right)
            if abs(l - r) > 1 or l == -1 or r == -1:
                return -1
            else:
                return max(l, r) + 1

    def isBalanced(self, root: TreeNode) -> bool:
        if not root:
            return True
        return self.dfs(root) > 0



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