[LeetCode] 107. Binary Tree Level Order Traversal II

轉自LeetCode

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]
]

<Solution>

這題是 Binary Tree Level Order Traversal 的衍生題

這次的 level 不是由上到下，而是要改成由下到上

可以完全用同樣觀念，只要改一行 code

就是將 level 的值丟到 ans 時，不要使用 push_back

而是改成 ans.insert(ans.begin(), out)，把高 level 的值放到低 level 的值前面

code 如下

	/**
	* Definition for a binary tree node.
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	public:
	vector<vector<int>> levelOrderBottom(TreeNode* root) {
	vector<vector<int>> ans;
	queue<TreeNode *> nodeQue;
	if(root) {
	nodeQue.push(root);
	}
	int size;
	vector<int> out;
	while(!nodeQue.empty()) {
	out.clear();
	size = nodeQue.size();
	for(int i = 0; i < size; i++) {
	TreeNode *tmpNode = nodeQue.front();
	nodeQue.pop();
	out.push_back(tmpNode->val);
	if(tmpNode->left) {
	nodeQue.push(tmpNode->left);
	}
	if(tmpNode->right) {
	nodeQue.push(tmpNode->right);
	}
	}
	//> put higher level into front
	ans.insert(ans.begin(),out);
	}
	return ans;
	}
	};

view raw binaryTreeLevelOrderTraversalII_1.cpp hosted with ❤ by GitHub

也可以不用使用 vector::insert() 的寫法，可以使用 std::reverse() 來反轉最後的答案就好

而且速度還比使用 vector::insert() 會快很多

code如下

	/**
	* Definition for a binary tree node.
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	public:
	vector<vector<int>> levelOrderBottom(TreeNode* root) {
	vector<vector<int>> ans;
	queue<TreeNode *> nodeQue;
	if(root) {
	nodeQue.push(root);
	}
	int size;
	vector<int> out;
	while(!nodeQue.empty()) {
	out.clear();
	size = nodeQue.size();
	for(int i = 0; i < size; i++) {
	TreeNode *tmpNode = nodeQue.front();
	nodeQue.pop();
	out.push_back(tmpNode->val);
	if(tmpNode->left) {
	nodeQue.push(tmpNode->left);
	}
	if(tmpNode->right) {
	nodeQue.push(tmpNode->right);
	}
	}
	ans.push_back(out);
	}
	//> reverse the ans
	reverse(ans.begin(), ans.end());
	return ans;
	}
	};

view raw binaryTreeLevelOrderTraversalII_3.cpp hosted with ❤ by GitHub

recursive 的寫法也是一樣，只要改成把高 level 的值放到低 level 的值前面就可以了

code 如下

	/**
	* Definition for a binary tree node.
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	private:
	vector<vector<int>> ans;
	public:
	vector<vector<int>> levelOrderBottom(TreeNode* root) {
	recursive(root, 0);
	return ans;
	}
	void recursive(TreeNode *node, const int &level) {
	if(!node) {
	return;
	}
	//> current node is in new level
	if(ans.size() == level) {
	//> put higher level into front
	ans.insert(ans.begin(), vector<int>());
	}
	ans[ans.size() - 1 - level].push_back(node->val);
	recursive(node->left, level+1);
	recursive(node->right, level+1);
	}
	};

view raw binaryTreeLevelOrderTraversalII_2.cpp hosted with ❤ by GitHub

也可使用 std::reverse() 就好

code 如下

	/**
	* Definition for a binary tree node.
	* struct TreeNode {
	* int val;
	* TreeNode *left;
	* TreeNode *right;
	* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
	* };
	*/
	class Solution {
	private:
	vector<vector<int>> ans;
	public:
	vector<vector<int>> levelOrderBottom(TreeNode* root) {
	recursive(root, 0);
	reverse(ans.begin(), ans.end());
	return ans;
	}
	void recursive(TreeNode *node, const int &level) {
	if(!node) {
	return;
	}
	//> current node is in new level
	if(ans.size() == level) {
	ans.push_back({});
	}
	ans[level].push_back(node->val);
	recursive(node->left, level+1);
	recursive(node->right, level+1);
	}
	};

view raw binaryTreeLevelOrderTraversalII_4.cpp hosted with ❤ by GitHub