102. Binary Tree Level Order Traversal
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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]]
题意:
给定一个二叉树,返回层次遍历的结果。结果为一个二维数组,每层的结点保存在一个数组中。
思路:
同BFS实现层次遍历,关键在于如何分层处理——在while(!q.empty())内加一层循环,以q的size为循环长度,每次循环处理删除一个q中的结点。这样在内存循环结束时,q中保存的就都是下一层的结点。
8ms
class Solution {public: vector<vector<int> > levelOrder(TreeNode *root) { vector<vector<int> > res; if (root == NULL) return res; queue<TreeNode*> q; q.push(root); while (!q.empty()) { vector<int> oneLevel; //保存每层的结果 int size = q.size(); for (int i = 0; i < size; ++i) { TreeNode *node = q.front(); q.pop(); oneLevel.push_back(node->val); if (node->left) q.push(node->left); if (node->right) q.push(node->right); } res.push_back(oneLevel); } return res; }};
此外,这道题还有递归的解法。核心在于需要一个二维数组,和一个变量level,当level递归到上一层的个数,新建一个空层,继续往里面加数字。可以想象为从左到右按列依次填充res的内容:
8ms
class Solution {public: vector<vector<int>> levelOrder(TreeNode* root) { vector<vector<int> > res; levelorder(root, 0, res); return res; } void levelorder(TreeNode *root, int level, vector<vector<int> > &res) { if (!root) return; if (res.size() == level) res.push_back({}); //如果递归的层数达到level,则在res中新建一个空层 res[level].push_back(root->val); if (root->left) levelorder(root->left, level + 1, res); if (root->right) levelorder(root->right, level + 1, res); }};
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