leetcode 之Binary Tree Level Order Traversal I和II 解题思路

题目如下：

Binary Tree Level Order Traversal I

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

解题思路：

层次遍历的时候需要借助辅助数据结构--队列。每一层用null节点区分一下。当删除队列的头节点时，需要判断当前节点是否为null,如果为null，说明该层已经结束，需要新建一个arraylist，同时需要再队列尾部添加null，用来区分下一层。但是需要注意的是，如果当前null为最后一个节点时，就不需要再在尾部添加null了，否则就会陷入死循环了。

代码如下：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List> levelOrder(TreeNode root) {
        List> list = new ArrayList>();
        if(root == null) return list;
        LinkedList queue = new LinkedList();
        queue.add(root);
        queue.add(null);
        List sublist = new ArrayList();
        while(queue.size() > 0){
            TreeNode node = queue.remove();
            if(node != null){
                sublist.add(node.val);
                if(node.left != null) queue.add(node.left);
                if(node.right != null) queue.add(node.right);
            }else{
                list.add(sublist);
                if(queue.size() > 0){
                    sublist = new ArrayList();
                    queue.add(null);
                }
                
            }
        }
        return list;
        
    }
}

Binary Tree Level Order Traversal II

 

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]
]

解法与I相同，只不过将最后结果翻转一下即可。这里的做法是，采用头插法。

代码如下：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List> levelOrderBottom(TreeNode root) {
        List> list = new ArrayList>();
        if(root == null) return list;
        LinkedList queue = new LinkedList();
        queue.add(root);
        queue.add(null);
        List sublist = new ArrayList();
        while(queue.size() > 0){
            TreeNode node = queue.remove();
            if(node != null){
                sublist.add(node.val);
                if(node.left != null) queue.add(node.left);
                if(node.right != null) queue.add(node.right);
            }else{
                list.add(0,sublist);//和I相比只有这里不同

                if(queue.size() > 0){
                    sublist = new ArrayList();
                    queue.add(null);
                }
                
            }
        }
        return list;
    }
}