107. Binary Tree Level Order Traversal II
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
Jamboard
Solution
/**
* 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;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> levels = new ArrayList<>();
level(root, levels, 0);
return levels;
}
void level(TreeNode node, List<List<Integer>> levels, int k) {
if (node == null) return;
if (levels.size() == k) {
levels.add(0, new ArrayList<>());
}
level(node.left, levels, k + 1);
level(node.right, levels, k + 1);
levels.get(levels.size() - k - 1).add(node.val);
}
}