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Commit6755383

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binary tree level order traversal
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packageeasy;
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importclasses.TreeNode;
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importjava.util.ArrayList;
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importjava.util.LinkedList;
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importjava.util.List;
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importjava.util.Queue;
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/**102. Binary Tree Level Order Traversal
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Total Accepted: 117221
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Total Submissions: 341219
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Difficulty: Easy
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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).
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For example:
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Given binary tree [3,9,20,null,null,15,7],
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3
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/ \
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9 20
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/ \
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15 7
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return its level order traversal as:
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[
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[3],
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[9,20],
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[15,7]
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]
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*/
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publicclassBinaryTreeLevelOrderTraversal {
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//Cheers, easily made it AC'ed on the first submission now, practice does make perfect!
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publicList<List<Integer>>levelOrder(TreeNoderoot) {
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List<List<Integer>>result =newArrayList<List<Integer>>();
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if(root ==null)returnresult;
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Queue<TreeNode>q =newLinkedList<TreeNode>();
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q.offer(root);
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while(!q.isEmpty()){
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List<Integer>thisLevel =newArrayList<Integer>();
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intqSize =q.size();
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for(inti =0;i <qSize;i++){
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TreeNodecurr =q.poll();
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thisLevel.add(curr.val);
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if(curr.left !=null)q.offer(curr.left);
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if(curr.right !=null)q.offer(curr.right);
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}
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result.add(thisLevel);
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}
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returnresult;
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}
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}

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