LeetCode | Binary Tree Level Order Traversal II

题目

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],
]

在Binary Tree Level Order Traversal基础上稍作修改。

代码

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;

public class BinaryTreeLevelOrderTraversalII {
	public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
		LinkedList<ArrayList<Integer>> results = new LinkedList<ArrayList<Integer>>();
		Queue<TreeNode> curLevel = new LinkedList<TreeNode>();
		Queue<TreeNode> nextLevel = new LinkedList<TreeNode>();
		if (root == null) {
			return new ArrayList<ArrayList<Integer>>();
		}
		curLevel.add(root);
		ArrayList<Integer> list = new ArrayList<Integer>();
		while (!curLevel.isEmpty()) {
			TreeNode node = curLevel.poll();
			list.add(node.val);
			if (node.left != null) {
				nextLevel.add(node.left);
			}
			if (node.right != null) {
				nextLevel.add(node.right);
			}
			if (curLevel.isEmpty()) {
				Queue<TreeNode> temp = curLevel;
				curLevel = nextLevel;
				nextLevel = temp;
				results.addFirst(list);
				list = new ArrayList<Integer>();
			}
		}

		return new ArrayList<ArrayList<Integer>>(results);
	}
}

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LeetCode | Binary Tree Level Order Traversal II

原文：http://blog.csdn.net/perfect8886/article/details/21108867