110.平衡二叉树

平衡二叉树需要看左右节点的深度是不是相差<=1

未看题解前

class Solution:def isBalanced(self, root: Optional[TreeNode]) -> bool:if not root:return Trueif abs(self.depth(root.left) - self.depth(root.right)) > 1:return Falsereturn self.isBalanced(root.left) and self.isBalanced(root.right)def depth(self, node):if not node:return 0return 1 + max(self.depth(node.left), self.depth(node.right))

用-1标记已经不满足平衡二叉树，最后结果不为-1说明为平衡二叉树。

class Solution:def isBalanced(self, root: Optional[TreeNode]) -> bool:# 求高度，后序遍历 左右中return self.getHeight(root) != -1def getHeight(self, node):if not node:return 0left_height = self.getHeight(node.left)if left_height == -1:return -1right_height = self.getHeight(node.right)if right_height == -1:return -1if abs(left_height - right_height) > 1:return -1return 1 + max(left_height, right_height)

257. 二叉树的所有路径

class Solution:def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:result = []if not root:return resultself.getPath(root, [root.val], result)return resultdef getPath(self, node, path, result):if not node.left and not node.right:result.append("->".join([str(i) for i in path]))if node.left:path.append(node.left.val)self.getPath(node.left, path, result)path.pop()if node.right:path.append(node.right.val)self.getPath(node.right, path, result)path.pop()

下面的代码更简洁一点

class Solution:def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:result = []if not root:return resultself.getPath(root, [], result)return resultdef getPath(self, node, path, result):path.append(node.val)if not node.left and not node.right:result.append("->".join([str(i) for i in path]))returnif node.left:self.getPath(node.left, path, result)path.pop()if node.right:self.getPath(node.right, path, result)path.pop()

404.左叶子之和

左叶子为左节点且为叶子节点

class Solution:def __init__(self):self.result = 0def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:if not root:return 0if root.left:if not root.left.left and not root.left.right:self.result += root.left.valelse:self.sumOfLeftLeaves(root.left)if root.right:self.sumOfLeftLeaves(root.right)return self.result

看了题解之后，代码更加简洁易懂

class Solution:def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:if not root:return 0left_leaves = self.sumOfLeftLeaves(root.left)if root.left and not root.left.left and not root.left.right:left_leaves = root.left.valright_leaves = self.sumOfLeftLeaves(root.right)return left_leaves + right_leaves