题目:
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]
]
confused what "{1,#,2,3}"
means? > read more on how binary tree is serialized on OJ.
链接: http://leetcode.com/problems/binary-tree-level-order-traversal/
题解:
层序遍历二叉树。使用BFS, 用一个Queue来辅助存储当前层的节点。
Time Complexity – O(n), Space Complexity – O(n) (最多一层节点数)
public class Solution {
public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();
if(root == null)
return result;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
ArrayList<Integer> list = new ArrayList<Integer>();
queue.add(root);
int curLevel = 1, nextLevel = 0; while(!queue.isEmpty()){
TreeNode node = queue.poll();
curLevel --;
list.add(node.val); if(node.left != null){
queue.offer(node.left);
nextLevel ++;
}
if(node.right != null){
queue.offer(node.right);
nextLevel ++;
}
if(curLevel == 0){
curLevel = nextLevel;
nextLevel = 0;
result.add(new ArrayList<Integer>(list));
list.clear();
}
} return result;
}
}
Update:
/**
* 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) {
List<List<Integer>> res = new ArrayList<>();
if(root == null)
return res;
Queue<TreeNode> queue = new LinkedList<>(); //BFS
ArrayList<Integer> list = new ArrayList<>();
queue.add(root); //Initialize
int curLevel = 1, nextLevel = 0; while(!queue.isEmpty()) {
TreeNode node = queue.poll();
list.add(node.val);
curLevel --;
if(node.left != null) {
queue.offer(node.left);
nextLevel++;
}
if(node.right != null) {
queue.offer(node.right);
nextLevel++;
}
if(curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(new ArrayList<Integer>(list));
list.clear();
}
} return res;
}
}
二刷:
使用一个queue来帮助BFS层序遍历,有两个记录当前层节点和下一层节点数的变量curLevel以及nextLevel。
Java:
/**
* 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) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0;
List<Integer> list = new ArrayList<>();
while (!q.isEmpty()) {
TreeNode node = q.poll();
curLevel--;
list.add(node.val);
if (node.left != null) {
q.offer(node.left);
nextLevel++;
}
if (node.right != null) {
q.offer(node.right);
nextLevel++;
}
if (curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(new ArrayList<Integer>(list));
list.clear();
}
}
return res;
}
}
三刷:
不要忘记把结果保存到结果集里,并且也要clear当前层。
Java:
/**
* 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) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) {
return res;
}
List<Integer> level = new ArrayList<>();
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1, nextLevel = 0; while (!q.isEmpty()) {
TreeNode node = q.poll();
curLevel--;
level.add(node.val);
if (node.left != null) {
q.offer(node.left);
nextLevel++;
}
if (node.right != null) {
q.offer(node.right);
nextLevel++;
}
if (curLevel == 0) {
curLevel = nextLevel;
nextLevel = 0;
res.add(new ArrayList<Integer>(level));
level.clear();
}
}
return res;
}
}
Update: 换种写法
/**
* 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) {
List<List<Integer>> res = new ArrayList<>();
if (root == null) return res;
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
int curLevel = 1;
List<Integer> list = new ArrayList<>(); while (!q.isEmpty()) {
TreeNode node = q.poll();
curLevel--;
list.add(node.val);
if (node.left != null) q.offer(node.left);
if (node.right != null) q.offer(node.right);
if (curLevel == 0) {
curLevel = q.size();
res.add(new ArrayList<>(list));
list.clear();
}
} return res;
}
}
Reference:
https://leetcode.com/discuss/21778/java-solution-using-dfs
https://leetcode.com/discuss/22533/java-solution-with-a-queue-used
https://leetcode.com/discuss/58454/java-queue-solution-beats-100%25