Binary Tree Level Order Traversal I和II 层次遍历二叉树

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its level order traversal as:

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

层次遍历二叉树，从左到右，从上到下

思路：利用两个队列分别保存当前这一层的节点和下一层的节点。当遍历某层节点时，插入下一层要遍历的节点

/** * 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>> levelOrder(TreeNode root) {        //存放结果        List<List<Integer>> res = new ArrayList<List<Integer>>();        if (root == null) {return res;}        //记录当前这一层的所有节点        LinkedList<TreeNode> currentLevel = new LinkedList<TreeNode>();        currentLevel.add(root);        while (currentLevel.size() > 0) {        //存储每层节点的值        List<Integer> currList = new ArrayList<Integer>();//存储下一层所有的节点LinkedList<TreeNode> nextLevel = new LinkedList<TreeNode>();while (currentLevel.size() > 0) {TreeNode node = currentLevel.poll();//如果有左右节点，那么加入到下一层队列中if (node.left != null) {nextLevel.add(node.left);}if (node.right != null) {nextLevel.add(node.right);}currList.add(node.val);}res.add(currList);currentLevel = nextLevel;        }                return res;    }}

第二个问题是：从底层向上遍历。

再把结果添加到结果集时，把上面的res.add(currList);改为res.add(0,currList);即每次都从下标为0的位置开始插入。