题干
binary-tree-level-order-traversal
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]
]binary-tree-level-order-traversal ii
Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree{3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its bottom-up level order traversal as:
[
[15,7]
[9,20],
[3],
]binary-tree-zigzag-level-order-traversal
Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree{3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]问题一：层序遍历。
问题二：层序遍历，但是要从下往上输出。
问题三：层序遍历，但是是zigzag形输出。
数据结构
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNodhttps://www.nowcoder.com/practice/d8566e765c8142b78438c133822b5118?tpId=46&tqId=29071&tPage=3&rp=3&ru=/ta/leetcode&qru=/ta/leetcode/question-ranking#e *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/解题思路
问题一
层序遍历，利用队列维护，每次把当前队列中的结点子结点送入队列，然后弹出当前结点，每一组为一层。
问题二
根据问题一只需加入reverse反转即可。
问题三
zigzag形状输出，则只需要加一个标志位来标示本层是从后向前遍历还是从前向后，从后向前reverse就可以。
参考代码
问题一：
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode *root) {
vector<vector<int>>res;
if (root==NULL)
return res;
queue<TreeNode *>data;//维护结点队列
data.push(root);
while(!data.empty())
{
int size=data.size();
vector<int>cur;//存储当前层
for(int i=1;i<=size;i++)
{
TreeNode *curnode=data.front();
data.pop();
cur.push_back(curnode->val);
if (curnode->left!=NULL) data.push(curnode->left);//存储当前结点的左右孩子结点
if (curnode->right!=NULL) data.push(curnode->right);
}
res.push_back(cur);
}
return res;
}
};问题二：
class Solution {
public:
vector<vector<int> > levelOrderBottom(TreeNode *root) {
vector<vector<int>>res;
if (root==NULL)
return res;
queue<TreeNode *>data;
data.push(root);
while(!data.empty())
{
int size=data.size();
vector<int>cur;
for(int i=1;i<=size;i++)
{
TreeNode *curnode=data.front();
data.pop();
cur.push_back(curnode->val);
if (curnode->left!=NULL) data.push(curnode->left);
if (curnode->right!=NULL) data.push(curnode->right);
}
res.push_back(cur);
}
reverse(res.begin(), res.end());//反转上层和底层序列
return res;
}
};问题三：
class Solution {
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
vector<vector<int>>res;
if (root==NULL)
return res;
queue<TreeNode *>data;
data.push(root);
int flag=0;//标志位
while(!data.empty())
{
int size=data.size();
vector<int>cur;
for(int i=1;i<=size;i++)
{
TreeNode *curnode=data.front();
data.pop();
cur.push_back(curnode->val);
if (curnode->left!=NULL) data.push(curnode->left);
if (curnode->right!=NULL) data.push(curnode->right);
}
if (flag%2==1)//根据标志位来判断是否要反转来实现zigzag
reverse(cur.begin(),cur.end());
flag++;
res.push_back(cur);
}
return res;
}
};方法讨论
三个题类似，都是树的层序遍历，主要是用队列维护，然后记录每层的结点数量进行循环就好。
易错点
reverse的使用。