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。