Binary Tree Level Order Traversal II | Leetcode

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.

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Solution: Add each level of nodes as a list. Collect all the children nodes first. Then collect the integer values as a list. Add this list to the front of the answer list.

/**
 * 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) {
        LinkedList<List<Integer>> answer = new LinkedList<>();
        if (root == null) return answer;
        List<TreeNode> nodes = new ArrayList<>();
        List<TreeNode> children = new ArrayList<>();
        nodes.add(root);
        do {
            List<Integer> list = new ArrayList<>();
            for (TreeNode node: nodes) {
                list.add(node.val);
                if (node.left != null) children.add(node.left);
                if (node.right != null) children.add(node.right);
            }
            answer.addFirst(list);
            nodes.clear();
            List<TreeNode> tmp = nodes;
            nodes = children;
            children = tmp;
        } while (!nodes.isEmpty());
        return answer;
    }
}