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;
}
}