【Lintcode】069.Binary Tree Level Order Traversal
阿新 • • 發佈:2017-05-10
vector pub i++ pre oot order ptr values logs
題目:
Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
ExampleGiven binary tree {3,9,20,#,#,15,7}
,
3 / 9 20 / 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
題解:
三種處理方法:queue + queue; queue + ‘dummy’;queue
Solution 1 ()
class Solution { /** * @param root: The root of binary tree. * @return: Level order a list of lists of integer */ public: vector<vector<int>> levelOrder(TreeNode *root) { vector<vector<int>> res; if (root == NULL) {return res; } queue<TreeNode*> queue; vector<int> levelNode; queue.push(root); queue.push(nullptr); while(!queue.empty()) { TreeNode *node = queue.front(); queue.pop(); if (node == nullptr) { res.push_back(levelNode); levelNode.resize(0); if (q.size() > 0) { q.push(nullptr); } } else { levelNode.push_back(root->val); if (node->left != nullptr) { queue.push(node->left); } if (node->right != nullptr) { queue.push(node->right); } } } return res; } };
Solution 2 ()
class Solution { /** * @param root: The root of binary tree. * @return: Level order a list of lists of integer */ public: vector<vector<int>> levelOrder(TreeNode *root) { vector<vector<int>> res; if (root == NULL) { return res; } queue<TreeNode*> queue; queue.push(root); while(!queue.empty()) { int size = queue.size(); vector<int> levelNode; for (int i = 0; i < size; i++) { TreeNode *node = queue.front(); queue.pop(); levelNode.push_back(node->val); if (node->left != nullptr) { queue.push(node->left); } if (node->right != nullptr) { queue.push(node->right); } } res.push_back(levelNode); } return res; } };
程序中的size是必要的,因為在for循環中queue是不斷變化的,那麽queue.size()也是不斷變化的,這就違背了我們的原則,這個size就是該層所有的非空node的個數,取完之後就要壓入result中,讀取下一層的節點
【Lintcode】069.Binary Tree Level Order Traversal