# Binary Tree Level Order Traversal II

### 描述​

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],]

### 递归版​

// Binary Tree Level Order Traversal II// 递归版，时间复杂度O(n)，空间复杂度O(n)public class Solution {    public List<List<Integer>> levelOrderBottom(TreeNode root) {        List<List<Integer>> result = new ArrayList<>();        traverse(root, 1, result);        Collections.reverse(result);        return result;    }    void traverse(TreeNode root, int level,                  List<List<Integer>> result) {        if (root == null) return;        if (level > result.size())            result.add(new ArrayList<>());        result.get(level-1).add(root.val);        traverse(root.left, level+1, result);        traverse(root.right, level+1, result);    }}

### 迭代版​

// Binary Tree Level Order Traversal II// 迭代版，时间复杂度O(n)，空间复杂度O(1)public class Solution {    public List<List<Integer>> levelOrderBottom(TreeNode root) {        List<List<Integer>> result = new ArrayList<>();        Queue<TreeNode> current = new LinkedList<>();        Queue<TreeNode> next = new LinkedList<>();        if(root == null) {            return result;        } else {            current.offer(root);        }        while (!current.isEmpty()) {            ArrayList<Integer> level = new ArrayList<>(); // elments in one level            while (!current.isEmpty()) {                TreeNode node = current.poll();                level.add(node.val);                if (node.left != null) next.add(node.left);                if (node.right != null) next.add(node.right);            }            result.add(level);            // swap            Queue<TreeNode> tmp = current;            current = next;            next = tmp;        }        Collections.reverse(result);        return result;    }}