Binary Tree Level Order Traversal II

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],

  1.     3
  2.    / \
  3.   9  20
  4.     /  \
  5.    15   7

return its bottom-up level order traversal as:

  1. [
  2.   [15,7],
  3.   [9,20],
  4.   [3]
  5. ]

Solution:

  1. /**
  2.  * Definition for binary tree
  3.  * public class TreeNode {
  4.  *     int val;
  5.  *     TreeNode left;
  6.  *     TreeNode right;
  7.  *     TreeNode(int x) { val = x; }
  8.  * }
  9.  */
  10. public class Solution {
  11.   public List<List<Integer>> levelOrderBottom(TreeNode root) {
  12.     List<List<Integer>> res = new ArrayList<List<Integer>>();
  13.     List<Integer> list = new ArrayList<Integer>();
  15.     if (root == null) {
  16.       return res;
  17.     }
  19.     Queue<TreeNode> curr = new LinkedList<TreeNode>();
  20.     Queue<TreeNode> next = new LinkedList<TreeNode>();
  22.     curr.add(root);
  24.     while (!curr.isEmpty()) {
  25.       TreeNode node = curr.poll();
  26.       list.add(node.val);
  28.       if (node.left != null) {
  29.         next.add(node.left);
  30.       }
  32.       if (node.right != null) {
  33.         next.add(node.right);
  34.       }
  36.       if (curr.isEmpty()) {
  37.         res.add(0, list);
  38.         list = new ArrayList<Integer>();
  40.         curr = next;
  41.         next = new LinkedList<TreeNode>();
  42.       }
  43.     }
  45.     return res;
  46.   }
  47. }