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import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;
/**
* Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).
*
* For example:
* Given binary tree {3,9,20,#,#,15,7},
* 3
* / \
* 9 20
* / \
* 15 7
* return its bottom-up level order traversal as:
* [
* [15,7]
* [9,20],
* [3],
* ]
* confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
*
*/
public class BinaryTreeLevelOrderTraversalII {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> ret = new ArrayList<ArrayList<Integer>>();
if (root == null)
return ret;
ArrayList<Integer> level = new ArrayList<Integer>();
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.add(root);
int currentLevel = 1;
int nextLevel = 0;
while (!queue.isEmpty()) {
TreeNode node = queue.remove();
level.add(node.val);
currentLevel--;
if (node.left != null) {
queue.add(node.left);
nextLevel++;
}
if (node.right != null) {
queue.add(node.right);
nextLevel++;
}
if (currentLevel == 0) {
ret.add(level);
level = new ArrayList<Integer>();
currentLevel = nextLevel;
nextLevel = 0;
}
}
int i = 0, j = ret.size() - 1;
while (i < j) {
ArrayList<Integer> tmp = ret.get(i);
ret.set(i, ret.get(j));
ret.set(j, tmp);
i++;
j--;
}
return ret;
}
}
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