标签：des style blog class code c
本文为senlie原创，转载请保留此地址：http://blog.csdn.net/zhengsenlie
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] ]
confused what "{1,#,2,3}" means? >
read more on how binary tree is serialized on OJ.
题意：给定一棵二叉树，返回按层遍历的结果
思路1：bfs，定义一个新的struct，记录指针向节点的指针和每个节点所在的层
复杂度1：时间O(n)，空间O(n)
思路2：dfs
递归函数：
void levelOrder(TreeNode *root, int level, vector<vector<int> >&result)
表示把根为root的树按层存放在result中，其中level表示当前的层数
复杂度2：时间O(n)，空间O(n)
相关题目：
Binary Tree Level Order Traversal II
Binary Tree Zigzag Level Order Traversal
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
struct NodeWithLevel{
TreeNode *p;
int level;
NodeWithLevel(TreeNode *pp, int l):p(pp), level(l){}
};
//思路1
vector<vector<int> > levelOrder(TreeNode *root){
vector<vector<int> > result;
queue<NodeWithLevel > q;
q.push(NodeWithLevel(root, 0));
while(!q.empty()){
NodeWithLevel cur = q.front();q.pop();
TreeNode *t = cur.p;
if(t != NULL){
if(result.size() <= cur.level) {
vector<int> v;
v.push_back(t->val);
result.push_back(v);
}else result[cur.level].push_back(t->val);
q.push(NodeWithLevel(t->left, cur.level + 1));
q.push(NodeWithLevel(t->right, cur.level + 1));
}
}
return result;
}
};/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
struct NodeWithLevel{
TreeNode *p;
int level;
NodeWithLevel(TreeNode *pp, int l):p(pp), level(l){}
};
//思路2
void levelOrder(TreeNode *root, int level, vector<vector<int> >&result)
{
if(!root) return ;
if(result.size() <= level) {
vector<int> v;
v.push_back(root->val);
result.push_back(v);
}else result[level].push_back(root->val);
levelOrder(root->left, level + 1, result);
levelOrder(root->right, level + 1, result);
}
vector<vector<int> > levelOrder(TreeNode *root){
vector<vector<int> >result;
levelOrder(root, 0, result);
return result;
}
};Leetcode 树 Binary Tree Level Order Traversal,布布扣,bubuko.com
Leetcode 树 Binary Tree Level Order Traversal
标签：des style blog class code c
原文地址：http://blog.csdn.net/zhengsenlie/article/details/25727655
