leetcode 虐我篇之（十八）Binary Tree Level Order Traversal II_given a complete binary tree, print the level orde-优快云博客

本文链接：https://blog.youkuaiyun.com/u013592466/article/details/38823109

本文介绍了一种从叶到根的二叉树层次遍历算法实现，通过双队列交替存储每层节点的方式，最终得到逆序的层次遍历结果。

做完Binary Tree Level Order Traversal这道题，再来做现在这道题Binary Tree Level Order Traversal II，描述如下：

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.

有了前一道题的解题思路和程序，做这道题，简直就是信手拈来，直接将最后的结果翻转后返回就ok了。代码如下：

std::vector<std::vector<int> > levelOrderBottom(TreeNode *root)
{
	std::vector<std::vector<int> > result,resultReverse;
	std::vector<int> temp;
	std::queue<TreeNode *> treeQueue1,treeQueue2;

	if (!root)
	{	
		return result;
	}

	treeQueue1.push(root);

	TreeNode *node = NULL;

	while(!treeQueue1.empty() || !treeQueue2.empty())
	{
		if (!treeQueue1.empty())
		{
			node = treeQueue1.front();
			treeQueue1.pop();
			if (node->left)
			{
				treeQueue2.push(node->left);
			}
			if (node->right)
			{
				treeQueue2.push(node->right);
			}

			temp.push_back(node->val);

		}

		if (treeQueue1.empty())
		{
			//this means that this level is over
			result.push_back(temp);
			temp.clear();
			treeQueue1.swap(treeQueue2);	//swap two queue
		}	

	}

	resultReverse.insert(resultReverse.begin(),result.rbegin(),result.rend());

	return resultReverse;
}

当然了，还有其他方法了，赶着去吃饭，就先这样上交着，回来再想。