102. Binary Tree Level Order Traversal

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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]
]

BFS

广度遍历写法是用一个队列，依次将各行节点存进去，poll 出来时将 val 存到 list 时顺便添加其子节点到队列中。代码如下：

public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> list = new ArrayList<List<Integer>> ();        if(root == null) return list;        Queue<TreeNode> q = new LinkedList<TreeNode> ();        q.add(root);        while(!q.isEmpty()) {            int size = q.size();            ArrayList<Integer> temp = new ArrayList<Integer> ();            for(int i = 0; i < size; i++) {                TreeNode poll = q.poll();                if(poll.left != null)q.add(poll.left);                if(poll.right != null)q.add(poll.right);                temp.add(poll.val);            }            list.add(temp);        }        return list;    }}

DFS

深搜，创建一个函数，参数为所要返回的 list，节点和该节点所在的 level，当 list 没有该 level 的子 list 时，则创建。每次将 val 存到对应level 的子list。之后递归调用。代码如下：

public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {        List<List<Integer>> list = new ArrayList<List<Integer>> ();        addList(list, root, 0);        return list;    }    private void addList(List<List<Integer>> list, TreeNode node, int level) {        if(node == null) return;        if(list.size() == level) list.add(new ArrayList<Integer> ());        List<Integer> temp = list.get(level);        temp.add(node.val);        addList(list, node.left, level + 1);        addList(list, node.right, level + 1);    }}

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