110 平衡二叉树

class Solution:
    def isBalanced(self, root: Optional[TreeNode]) -> bool:
        def height(root):
            if not root:
                return 0
            return 1+max(height(root.left),height(root.right))
        if not root:
            return True
        return abs(height(root.left)-height(root.right))<=1 and self.isBalanced(root.left) and self.isBalanced(root.right)

class Solution:
    def isBalanced(self, root: Optional[TreeNode]) -> bool:

        def height(root):
            if not root:
                return 0
            l=height(root.left)
            r=height(root.right)
            if l==-1 or r==-1 or abs(l-r)>1:
                return -1
            return 1+max(l,r)
        return height(root)>=0

257 二叉树的所有路径

class Solution:
    def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]:
        def sub(root,path):
            if not root:
                return 
            path+=str(root.val)
            if not root.left and not root.right:
                paths.append(path)
                return
            path+="->"
            sub(root.left,path)
            sub(root.right,path)
        paths=[]
        sub(root,'')
        return paths

404 左叶子之和

class Solution:
    def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int:
        isLeafNode=lambda node: not node.left and not node.right
        def dfs(root):
            if not root:
                return 0
            ans=0
            if root.left:
                ans+=root.left.val if isLeafNode(root.left) else dfs(root.left)
            if root.right and not isLeafNode(root.right):
                ans+=dfs(root.right)
            return ans
        if not root:
            return 0
        return dfs(root)