Binary Tree Level Order Traversal
Binary Tree Level Order Traversal first search
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]
]
Binary Tree Level Order Traversal Solution
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class BinarySearchTreeLevelOrderTraversal {
public List<List<Integer>> levelOrder(TreeNode root) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(root == null) {
return result;
}
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
List<Integer> tempList = new ArrayList<Integer>();
tempList.add(root.val);
result.add(tempList);
while(!queue.isEmpty()) {
Queue<TreeNode> currentLevel = new LinkedList<TreeNode>();
List<Integer> list = new ArrayList<Integer>();
while(!queue.isEmpty()) {
TreeNode current = queue.remove();
if(current.left != null) {
currentLevel.add(current.left);
list.add(current.left.val);
}
if(current.right != null) {
currentLevel.add(current.right);
list.add(current.right.val);
}
}
if(list.size() > 0) {
result.add(list);
}
queue = currentLevel;
}
return result;
}
}
Last modified October 4, 2020