LeetCode-in-Java
107. Binary Tree Level Order Traversal II
Medium
Given the root of a binary tree, return the bottom-up level order traversal of its nodes’ values. (i.e., from left to right, level by level from leaf to root).
Example 1:
Input: root = [3,9,20,null,null,15,7]
Output: [[15,7],[9,20],[3]]
Example 2:
Input: root = [1]
Output: [[1]]
Example 3:
Input: root = []
Output: []
Constraints:
- The number of nodes in the tree is in the range
[0, 2000]. -1000 <= Node.val <= 1000
Solution
import com_github_leetcode.TreeNode;
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
/*
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
private List<List<Integer>> order = new ArrayList<>();
public List<List<Integer>> levelOrderBottom(TreeNode root) {
getOrder(root, 0);
Collections.reverse(order);
return order;
}
public void getOrder(TreeNode root, int level) {
if (root != null) {
if (level + 1 > order.size()) {
order.add(new ArrayList<>());
}
order.get(level).add(root.val);
getOrder(root.left, level + 1);
getOrder(root.right, level + 1);
}
}
}