给定一个二叉树,返回其节点值自底向上的层次遍历。 (即按从叶子节点所在层到根节点所在的层,逐层从左向右遍历)
例如: 给定二叉树 [3,9,20,null,null,15,7],
3 / \ 9 20 / \ 15 7
返回其自底向上的层次遍历为:
[ [15,7], [9,20], [3] ]
利用层次遍历,层次遍历的时候进入下一层的时候记录一下当前队列中有几个元素。
class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { List<List<Integer>> res = new LinkedList<>(); if (root == null) { return res; } LinkedList<TreeNode> queue = new LinkedList<>(); queue.add(root); while (!queue.isEmpty()) { int size = queue.size(); List<Integer> levelVal = new LinkedList<>(); while (size > 0) { TreeNode current = queue.poll(); if (current.left != null) { queue.add(current.left); } if (current.right != null) { queue.add(current.right); } levelVal.add(current.val); size--; } res.add(0, levelVal); } return res; } }
用递归去做。 用递归去做的关键在于需要把层数也带上。
class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { List<List<Integer>> res = new LinkedList<>(); if (root == null) { return res; } helper(root, res, 0); return res; } public void helper(TreeNode root, List<List<Integer>> res, int depth) { if (root == null) { return; } if (depth == res.size()) { res.add(0, new LinkedList<>()); } List<Integer> current = res.get(res.size() - depth - 1); helper(root.left, res, depth + 1); helper(root.right, res, depth + 1); current.add(root.val); } }
手撕代码 QQ 群:805423079, 群密码:1024
1 snowonion 2019-03-15 11:01:18 +08:00 https://www.codewars.com/kata/sort-binary-tree-by-levels 的 Haskell 解法的高票答案,稍加修改就能用在这里。 (剧透警告) 注意这里二叉树的定义方式是 ```haskell data TreeNode a = TreeNode { left :: Maybe (TreeNode a), right :: Maybe (TreeNode a), value :: a } deriving Show ``` 解答: ```haskell levelOrderBottomUpHierarchical :: Maybe (TreeNode a) -> [[a]] levelOrderBottomUpHierarchical = reverse . takeWhile (not . null) . go where go Nothing = repeat [] go (Just x) = [value x] : zipWith (++) (go $ left x) (go $ right x) ``` 测试: ```haskell leaf v = Just (TreeNode {left = Nothing, right = Nothing, value = v}) t3 = Just (TreeNode { left = leaf 9, right = Just (TreeNode { left = leaf 15, right = leaf 7, value = 20 }), value = 3 }) -- ghci 执行 -- > levelOrderBottomUpHierarchical t3 -- [[15,7],[9,20],[3]] ``` |