python3 解法, 执行用时: 52 ms, 内存消耗: 18.5 MB, 提交时间: 2020-11-18 17:04:45

# 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:
        # 计算节点高度
        def height(root: TreeNode):
            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
            return max(leftHeight, rightHeight) + 1
            
        return height(root) >= 0

golang 解法, 执行用时: 4 ms, 内存消耗: 6 MB, 提交时间: 2020-11-18 16:58:43

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func isBalanced(root *TreeNode) bool {
    if root == nil {
        return true
    }
    return math.Abs(float64(height(root.Left) - height(root.Right))) <= 1  && isBalanced(root.Left) && isBalanced(root.Right)
}

func height(node *TreeNode) int {
    if node == nil {
        return 0
    }
    return max(height(node.Left), height(node.Right)) + 1
}

func max(a, b int) int {
    if a > b {
        return a
    }
    return b
}

python3 解法, 执行用时: 84 ms, 内存消耗: 20.4 MB, 提交时间: 2020-11-18 16:52:18

# 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:
        # 计算节点高度
        def height(root: TreeNode):
            if not root:
                return 0
            return max(height(root.left), height(root.right)) + 1
            
        if not root:
            return True
        return self.isBalanced(root.left) and self.isBalanced(root.right) and abs(height(root.left) - height(root.right)) <= 1