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Solution.java
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Solution.java
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package binarylevelordertraversal;
import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.List;
/**
* 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]
* ]
*/
class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
class Solution {
List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> results = new ArrayList<>();
if(root == null) return results;
ArrayDeque<TreeNode> ad = new ArrayDeque<>();
// BFS - add root to queue and begin
ad.add(root);
TreeNode current;
List<Integer> level;
int currLevelLen = -1;
// run until queue empty
while(!ad.isEmpty()) {
// begin current level
level = new ArrayList<>();
// run loop for as many children there are in queue
currLevelLen = ad.size();
for(int i = 0; i < currLevelLen; i++) {
// get front of queue
current = ad.pollFirst();
if(current != null) {
// if a valid node, get its children, add current to level and
// add children to queue, if not null
TreeNode left = current.left;
TreeNode right = current.right;
level.add(current.val);
if(left != null) {
ad.add(left);
}
if(right != null) {
ad.add(right);
}
}
}
// ADD current level to results only after this particular level is over
// which will be when for loop (= size of queue) finishes
if(!level.isEmpty()) {
results.add(level);
}
}
return results;
}
}
/*
Success
Details
Runtime: 1 ms, faster than 72.47% of Java online submissions for Binary Tree Level Order Traversal.
Memory Usage: 39 MB, less than 5.33% of Java online submissions for Binary Tree Level Order Traversal.
*/