LeetCode - Binary Tree Level Order Traversal
Problem description
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
For example:
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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]
]
Analysis
A simple BFS problem.
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/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
Queue<TreeNode> q = new ArrayDeque<>();
List<List<Integer>> res = new ArrayList<>();
if (root == null){
return res;
}
q.offer(root);
while (!q.isEmpty()){
List<Integer> list = new ArrayList<>();
List<TreeNode> next = new ArrayList<>();
while(!q.isEmpty()){
TreeNode node = q.poll();
list.add(node.val);
if (node.left != null){
next.add(node.left);
}
if (node.right != null){
next.add(node.right);
}
}
res.add(list);
for (TreeNode node:next){
q.offer(node);
}
}
return res;
}
}
In each iteration, we don’t need to use another array to store the next level. However, we can use q.size() to find out how many item this level will have.
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class Solution {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> levels = new ArrayList<List<Integer>>();
if (root == null) return levels;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
int level = 0;
while ( !queue.isEmpty() ) {
// start the current level
levels.add(new ArrayList<Integer>());
// number of elements in the current level
int level_length = queue.size();
for(int i = 0; i < level_length; ++i) {
TreeNode node = queue.remove();
// fulfill the current level
levels.get(level).add(node.val);
// add child nodes of the current level
// in the queue for the next level
if (node.left != null) queue.add(node.left);
if (node.right != null) queue.add(node.right);
}
// go to next level
level++;
}
return levels;
}
}
What to improve
- standard BFS use q.size() for iteration.