107.Binary Tree Level Order Traversal II

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Solution 1: accepted 2ms

BFS. Simply reverse the result from question 102. 

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public List<List<Integer>> levelOrderBottom(TreeNode root) {
    List<List<Integer>> result = new ArrayList<List<Integer>>();
    if (root == null) {
      return result;
    }
    Queue<TreeNode> queue = new LinkedList<>();
    queue.add(root);
    while (!queue.isEmpty()) {
      ArrayList<Integer> list = new ArrayList<>();
      int size = queue.size();
      for (int i = 0; i < size; i++) {
        TreeNode node = queue.poll();
        list.add(node.val);
        if (node.left != null) {
          queue.add(node.left);
        }
        if (node.right != null) {
          queue.add(node.right);
        }
      }
      result.add(list);
    }
    Collections.reverse(result);
    return result;
}