leetcode107. 二叉树的层序遍历 II

1.题目描述：

给你二叉树的根节点root，返回其节点值自底向上的层序遍历。（即按从叶子节点所在层到根节点所在的层，逐层从左向右遍历）

 2.层序遍历：普通层序遍历再将集合反转。

/**
 * 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>> resList = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        if (root == null) return resList;
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            while (size > 0) {
                TreeNode temp = queue.poll();
                list.add(temp.val);
                size--;
                if (temp.left != null) queue.offer(temp.left);
                if (temp.right != null) queue.offer(temp.right);
            }
            resList.add(list);
        }
        Collections.reverse(resList);
        return resList;
    }
}

或： 

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> resList = new ArrayList<>();
        Queue<TreeNode> queue = new LinkedList<>();
        if (root == null) return resList;
        queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            List<Integer> list = new ArrayList<>();
            while (size > 0) {
                TreeNode temp = queue.poll();
                list.add(temp.val);
                size--;
                if (temp.left != null) queue.offer(temp.left);
                if (temp.right != null) queue.offer(temp.right);
            }
            resList.add(0, list);//头插
        }
        return resList;
    }
}

3.递归，深度优先：

class Solution {
    private List<List<Integer>> list = new ArrayList<>();
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        if(root == null) return list;
        dfs(root, 1);
        return list;
    }

    public void dfs(TreeNode root, int depth) {
        if (root == null) return;
        if (depth > list.size()) list.add(0, new ArrayList<>());//新增行始终插在最前
        list.get(list.size() - depth).add(root.val);//倒着查找插入元素
        dfs(root.left, depth + 1); 
        dfs(root.right, depth + 1);
    }
}