Contents
Question
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,null,null,15,7],
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[ [15,7], [9,20], [3] ]
Solution
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> rlist = new LinkedList<>();
if (root == null) return rlist;
Queue<TreeNode> queue = new LinkedList<>();
queue.add(root);
while (!queue.isEmpty()) {
int size = queue.size();
List<Integer> ilist = new LinkedList<>();
for (int i = 0; i < size; i++) {
TreeNode curNode = queue.remove();
ilist.add(curNode.val);
if (curNode.left != null) {
queue.add(curNode.left);
}
if (curNode.right != null) {
queue.add(curNode.right);
}
}
rlist.add(ilist);
}
Collections.reverse(rlist);
return rlist;
}
}
Execution time: 3 ms , defeated in all Java commits 65.43% User
Memory consumption: 35.2 MB , defeated in all Java commits 94.34% User