107 Binary Tree Level Order Traversal II

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本文介绍了一种解决二叉树底部向上层次遍历的方法，提供了两种实现思路：深度优先搜索（DFS）的递归方式及广度优先搜索（BFS）的队列方式，并附带详细的代码实现。

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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],

    3
    / \
    9  20
    /  \
    15   7

    return its bottom-up level order traversal as:

    [
    [15,7],
    [9,20],
    [3]
    ]

    //DFS,递归，分方法，看不懂
   /* public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
        levelMaker(wrapList, root, 0);
        return wrapList;
    }

    public void levelMaker(List<List<Integer>> list, TreeNode root, int level) {
        if (root == null) return;
        if (level >= list.size()) {
            list.add(0, new LinkedList<Integer>());
        }
        levelMaker(list, root.left, level + 1);//level不变哦
        levelMaker(list, root.right, level + 1);
        list.get(list.size() - level - 1).add(root.val);
    }*/


    //BFS，队列，链表
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> wrapList = new LinkedList<List<Integer>>();
        Queue<TreeNode> queue = new LinkedList<>();
        if (root == null) return wrapList;

        queue.offer(root);
        while (!queue.isEmpty()) {
            int width = queue.size();
            List<Integer> subList = new LinkedList<>();
            for (int i = 0; i < width; i++) {//queue.size（）不能放在这！！！！因为过程中queue会变！！！！导致上限变化
                if (queue.peek().left != null) queue.offer(queue.peek().left);
                if (queue.peek().right != null) queue.offer(queue.peek().right);
                subList.add(queue.poll().val);
            }
            wrapList.add(0, subList);
        }
        return wrapList;
    }

    public class TreeNode {
        int val;
        TreeNode left;
        TreeNode right;

        TreeNode(int x) {
            val = x;
        }
    }