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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,#,#,15,7},
3
/ 9 20
/ 15 7
return its level order traversal as:
[ [3], [9,20], [15,7] ]
层序遍历，既可以用dfs实现也可以用bfs实现，用bfs实现的时候应该借助队列，代码如下：
dfs:
1 class Solution { 2 public: 3 vector<vector<int>> levelOrder(TreeNode* root) { 4 dfs(root, 0); 5 return ret; 6 } 7 void dfs(TreeNode * root, int dep) 8 { 9 if(!root) return; 10 if(dep < ret.size()){ 11 ret[dep].push_back(root->val); 12 }else{ 13 vector<int> tmp; 14 tmp.push_back(root->val); 15 ret.push_back(tmp); 16 } 17 dfs(root->left, dep+1); 18 dfs(root->right, dep+1); 19 } 20 private: 21 vector<vector<int>> ret; 22 };
bfs:
1 class Solution { 2 private: 3 struct Node{ 4 TreeNode * treeNode; 5 int level; 6 Node(){} 7 Node(TreeNode * node, int lv) 8 :treeNode(node), level(lv){} 9 }; 10 public: 11 vector<vector<int>> levelOrder(TreeNode* root) { 12 queue<Node> nodeQueue; 13 vector<vector<int>> ret; 14 if(root == NULL) 15 return ret; 16 nodeQueue.push(Node(root, 0)); 17 int dep = -1; 18 while(!nodeQueue.empty()){ 19 Node node = nodeQueue.front(); 20 if(node.treeNode->left) 21 nodeQueue.push(Node(node.treeNode->left, node.level + 1)); 22 if(node.treeNode->right) 23 nodeQueue.push(Node(node.treeNode->right, node.level + 1)); 24 if(dep == node.level) 25 ret[dep].push_back(node.treeNode->val); 26 else{ 27 vector<int> tmp; 28 dep++; 29 ret.push_back(tmp); 30 ret[dep].push_back(node.treeNode->val); 31 } 32 nodeQueue.pop(); 33 } 34 return ret; 35 } 36 };
LeetCode OJ：Binary Tree Level Order Traversal（二叉树的层序遍历）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/4905683.html
