[LeetCode] 102. Binary Tree Level Order Traversal

轉自LeetCode

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

<Solution>

這題是要把在同高度的 node 的值 group 在一起

想法如下

• 一層一層處理，把同一層的 node，按照從左至右的順序放到 queue

• 只要 queue 不為空，就代表還沒歷遍完

• 當下 queue 的 size，就代表有多少個 node 在同一層 level，逐一取出來，並將值放到答案

code 如下
c++

	/**
	* 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>> levelOrder(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();
	//> get all nodes in current level
	for(int i = 0; i < size; i++) {
	TreeNode *tmpNode = nodeQue.front();
	nodeQue.pop();
	out.push_back(tmpNode->val);
	//> if current node has left node
	if(tmpNode->left) {
	nodeQue.push(tmpNode->left);
	}
	//> if current node has right node
	if(tmpNode->right) {
	nodeQue.push(tmpNode->right);
	}
	}
	//> put all values of nodes of same level into ans
	ans.push_back(out);
	}
	return ans;
	}
	};

view raw binaryTreeLevelOrderTraversal_1.cpp hosted with ❤ by GitHub

kotlin

	/**
	* Example:
	* var ti = TreeNode(5)
	* var v = ti.`val`
	* Definition for a binary tree node.
	* class TreeNode(var `val`: Int) {
	* var left: TreeNode? = null
	* var right: TreeNode? = null
	* }
	*/
	class Solution {
	fun levelOrder(root: TreeNode?): List<List<Int>> {
	val ans = mutableListOf<List<Int>>()
	val queue = mutableListOf<TreeNode>()
	root?.run { queue.add(this) }
	while(queue.isNotEmpty()) {
	val size = queue.size
	val output = mutableListOf<Int>()
	for(i in 0 until size) {
	val tmp = queue.first()
	output.add(tmp.`val`)
	queue.removeAt(0)
	tmp.left?.let { queue.add(it) }
	tmp.right?.let { queue.add(it) }
	}
	ans.add(output)
	}
	return ans
	}
	}

view raw binary_tree_level_order_traversal.kt hosted with ❤ by GitHub

這邊再看一個 recursive 的寫法 (參考資料)

想法也是一樣，根據 level 去將每個 node 的值填到對應的 vector 上

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>> levelOrder(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) {
	ans.push_back({});
	}
	ans[level].push_back(node->val);
	recursive(node->left, level+1);
	recursive(node->right, level+1);
	}
	};

view raw binaryTreeLevelOrderTraversal_2.cpp hosted with ❤ by GitHub