Binary Tree Level Order Traversal II

Binary Tree Level Order Traversal II

问题：

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).

思路：

　　层次遍历 BFS

我的代码：

public class Solution {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();

if (root == null) {
return result;
}

Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
int currLevelNodeNum = 1;
int nextLevelNodeNum = 0;

while (currLevelNodeNum != 0) {
ArrayList<Integer> currLevelResult = new ArrayList<Integer>();
nextLevelNodeNum = 0;

while (currLevelNodeNum != 0) {
TreeNode node = queue.poll();

currLevelNodeNum--;
currLevelResult.add(node.val);

if (node.left != null) {
queue.offer(node.left);
nextLevelNodeNum++;
}

if (node.right != null) {
queue.offer(node.right);
nextLevelNodeNum++;
}
}

result.add(0, currLevelResult);
currLevelNodeNum = nextLevelNodeNum;
}
return result;
}
}

View Code
学习之处：

ArrayList竟然还有这个add(index,Object)功能，如此便省去了翻转链表的时间，插入表头的时间O(n)，翻转链表的时间O(n)，实际的程序运行时间也一致。