题目描述：
方法一：
# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]: if not root: return [] stack = [root] ans = [] while stack: tem_stack=[] tem_ans = [] for i in stack: tem_ans.append(i.val) if i.left: tem_stack.append(i.left) if i.right: tem_stack.append(i.right) stack = tem_stack ans.append(tem_ans) for i in range(len(ans)): if i%2!=0: ans[i] = ans[i][::-1] return ans
另：
# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]: if not root: return [] stack = [root] ans = [] depth = 0 while stack: tem_ans,tem_stack = [],[] for i in stack: tem_ans.append(i.val) if i.left: tem_stack.append(i.left) if i.right: tem_stack.append(i.right) if depth%2 == 0: ans.append(tem_ans) else: ans.append(tem_ans[::-1]) stack = tem_stack depth += 1 return ans
方法二：递归
# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def zigzagLevelOrder(self, root: TreeNode) -> List[List[int]]: res = [] def helper(root, depth): if not root: return if len(res) == depth: res.append([]) if depth % 2 == 0: res[depth].append(root.val) else: res[depth].insert(0, root.val) helper(root.left, depth + 1) helper(root.right, depth + 1) helper(root, 0) return res