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] ]
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
/** * 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) { List<List<Integer>> res = new ArrayList<List<Integer>>(); if (root == null) { return res; } LinkedList<TreeNode> linkedList = new LinkedList<TreeNode>(); linkedList.add(root); int cur = 1; int next = 0; ArrayList<Integer> arrayList = new ArrayList<Integer>(); while (!linkedList.isEmpty()) { TreeNode first = linkedList.poll(); cur--; arrayList.add(first.val); if (first.left != null) { linkedList.add(first.left); next++; } if (first.right != null) { linkedList.add(first.right); next++; } if (cur == 0) { cur = next; next = 0; res.add(0, arrayList); arrayList = new ArrayList<Integer>(); } } return res; } }
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