Binary Tree Level Order Traversal
Source
Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level). Example Given binary tree {3,9,20,#,#,15,7}, 3 / \ 9 20 / \ 15 7 return its level order traversal as: [ [3], [9,20], [15,7] ] Challenge Using only 1 queue to implement it.
题解 - 使用队列
此题为广搜的基础题,使用一个队列保存每层的节点即可。出队和将子节点入队的实现使用 for 循环,将每一轮的节点输出。
C++
/** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ 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> > result; if (NULL == root) { return result; } queue<TreeNode *> q; q.push(root); while (!q.empty()) { vector<int> list; int size = q.size(); // keep the queue size first for (int i = 0; i != size; ++i) { TreeNode * node = q.front(); q.pop(); list.push_back(node->val); if (node->left) { q.push(node->left); } if (node->right) { q.push(node->right); } } result.push_back(list); } return result; } };
Java
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { public List<List<Integer>> levelOrder(TreeNode root) { List<List<Integer>> result = new ArrayList<List<Integer>>(); if (root == null) return result; Queue<TreeNode> q = new LinkedList<TreeNode>(); q.offer(root); while (!q.isEmpty()) { List<Integer> list = new ArrayList<Integer>(); int qSize = q.size(); for (int i = 0; i < qSize; i++) { TreeNode node = q.poll(); list.add(node.val); // push child node into queue if (node.left != null) q.offer(node.left); if (node.right != null) q.offer(node.right); } result.add(new ArrayList<Integer>(list)); } return result; } }
源码分析
- 异常,还是异常
- 使用 STL 的 queue 数据结构,将 root 添加进队列
- 遍历当前层所有节点,注意需要先保存队列大小,因为在入队出队时队列大小会变化
- list 保存每层节点的值,每次使用均要初始化
复杂度分析
使用辅助队列,空间复杂度 O(n), 时间复杂度 O(n).