107. Binary Tree Level Order Traversal II

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107. 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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

[
  [15,7],
  [9,20],
  [3]
]

解析

• 题目虽然思路很清晰，但是查看了方法2,3感觉也吸收了很多思路！

// Binary Tree Level Order Traversal II
class Solution_107 {
public:
	//bfs
	vector<vector<int>> levelOrderBottom(TreeNode* root) {
		
		vector<vector<int> > vecs;
		stack<vector<int>> sta;
		if (!root)
		{
			return vecs;
		}

		queue<TreeNode*> que;
		que.push(root);

		while (!que.empty())
		{
			int size = que.size();
			vector<int> vec;
			TreeNode* temp;
			for (int i = 0; i < size;i++)
			{
				temp = que.front();
				que.pop();

				vec.push_back(temp->val);

				if (temp->left)
				{
					que.push(temp->left);
				}
				if (temp->right)
				{
					que.push(temp->right);
				}
			}
			sta.push(vec);
		}

		// reverse(res.begin(),res.end());
		while (!sta.empty())
		{
			vector<int> vec = sta.top();
			sta.pop();
			vecs.push_back(vec);
		}

		return vecs;
	}

    // 记录每层的个数；curCount当前层个数，nextCount下一层个节点个数
	vector<vector<int>> levelOrderBottom1(TreeNode* root) {
		int curCount = 0, nextCount = 0, level = 0;
		vector<vector<int>> ret;
		if (!root) return ret;

		curCount = 1;
		queue<TreeNode*> q;
		q.push(root);
		ret.push_back(vector<int>(0, curCount));

		while (!q.empty()) {
			if (curCount == 0) {
				curCount = nextCount;
				nextCount = 0;
				level++;
				ret.push_back(vector<int>(0, curCount)); //初始化下一层vector的大小
			}

			TreeNode *node = q.front();
			q.pop();
			curCount--;
			ret[level].push_back(node->val); //在同一层的节点加入

			if (node->left) {
				nextCount++;
				q.push(node->left);
			}
			if (node->right) {
				nextCount++;
				q.push(node->right);
			}
		}
		reverse(ret.begin(), ret.end());
		return ret;
	}


	//dfs
	int getHeight(TreeNode *root)
	{
		if (!root) 
			return 0;
		return max(getHeight(root->left), getHeight(root->right)) + 1;
	}
	vector<vector<int> > levelOrderBottom2(TreeNode *root)
	{
		if (!root) 
			return vector<vector<int>>();

		vector<vector<int>> res(getHeight(root), vector<int>());

		dfs(root, res.size() - 1, res);
		return res;
	}
	void dfs(TreeNode *root, int height, vector<vector<int>> &res)
	{
		if (!root)
			return;
		res[height].push_back(root->val);

		dfs(root->left, height - 1, res);
		dfs(root->right, height - 1, res);
	}
};

题目来源

• 107. Binary Tree Level Order Traversal II