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Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
同样是二叉树广度优先遍历的变形,要求从下面开始记录每一层的节点,我们可以从根节点开始,倒序记录每一层的节点就可以了。代码如下:
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7],
[9,20],
[3]
]
同样是二叉树广度优先遍历的变形,要求从下面开始记录每一层的节点,我们可以从根节点开始,倒序记录每一层的节点就可以了。代码如下:
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<List<Integer>> levelOrderBottom(TreeNode root) { LinkedList<List<Integer>> result = new LinkedList<List<Integer>>(); List<Integer> list = new ArrayList<Integer>(); Queue<TreeNode> queue = new LinkedList<TreeNode>(); if(root == null) return result; queue.offer(root); int helper = 1, count = 0; while(!queue.isEmpty()) { TreeNode cur = queue.poll(); list.add(cur.val); helper --; if(cur.left != null) { queue.offer(cur.left); count ++; } if(cur.right != null) { queue.offer(cur.right); count ++; } if(helper == 0) { result.addFirst(new ArrayList<Integer>(list)); helper = count; count = 0; list.clear(); } } return result; } }
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