[LeetCode] 199. Binary Tree Right Side View
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这题的意思就是人站在一棵二叉树右边
1 <--- / \2 3 <--- \ \ 5 4 <---先是自己瞎想提交了好几次都不对,然后网上看了看,很多人都是用的一层一层地遍历然后存入最右边的node.val
我的方法是类似遍历前序二叉树,每一层的深度通过deep记录下来然后更新cache
class Solution(object): def rightSideView(self, root): """ :type root: TreeNode :rtype: List[int] """ if not root: return [] self.cache = [] self.deep = 1 self.Read(root) return self.cache def Read(self,node): if self.deep > len(self.cache): self.cache.append(node.val) else : self.cache[self.deep - 1] = node.val if node.left is not None: self.deep += 1 self.Read(node.left) if node.right is not None: if node.left is None: self.deep += 1 self.Read(node.right) if node.left is None and node.right is None: self.deep += 1 self.deep -= 1
速度比较慢...
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