103. Binary Tree Zigzag Level Order Traversal
题目
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]
]
confused what"{1,#,2,3}"means? > read more on how binary tree is serialized on OJ.
OJ's Binary Tree Serialization:
The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below.
Here's an example:
1
/ 2 3
/
4
5
The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".
解析
// 103. Binary Tree Zigzag Level Order Traversal
class Solution_103 {
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
vector<vector<int> > vecs;
if (!root)
{
return vecs;
}
queue<TreeNode*> que;
que.push(root);
int level = 0;
while (!que.empty())
{
int size = que.size();
vector<int> vec;
while (size--)
{
TreeNode* temp = que.front();
que.pop();
vec.push_back(temp->val);
if (temp->left)
{
que.push(temp->left);
}
if (temp->right)
{
que.push(temp->right);
}
}
level++;
if (!(level % 2)) //实际是偶数行反转，从第0行开始 //奇数层反转vector bug:level % 2 != 0
{
reverse(vec.begin(), vec.end());
}
vecs.push_back(vec);
}
return vecs;
}
};
- 思路2：设置标志位，对奇数行取vector对应的元素
vector<vector<int> > zigzagLevelOrder(TreeNode* root) {
if (root == NULL) {
return vector<vector<int> > ();
}
vector<vector<int> > result;
queue<TreeNode*> nodesQueue;
nodesQueue.push(root);
bool leftToRight = true;
while ( !nodesQueue.empty()) {
int size = nodesQueue.size();
vector<int> row(size);
for (int i = 0; i < size; i++) {
TreeNode* node = nodesQueue.front();
nodesQueue.pop();
// find position to fill node's value
int index = (leftToRight) ? i : (size - 1 - i);
row[index] = node->val;
if (node->left) {
nodesQueue.push(node->left);
}
if (node->right) {
nodesQueue.push(node->right);
}
}
// after this level
leftToRight = !leftToRight;
result.push_back(row);
}
return result;
}
- 思路3：使用deque双端队列，注意取back()元素，先进right孩子
class Solution {
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
vector<vector<int> > ans;
if(root == NULL) return ans;
deque<TreeNode*> dq;
dq.push_back(root);
int zz = 0;
while(dq.size())
{
vector<int> tmp;
int h = dq.size();
int l = 0;
TreeNode *t;
while(l < h)
{
if(zz % 2 == 0)
{
t = dq.front();
dq.pop_front();
if(t->left) dq.push_back(t->left);
if(t->right) dq.push_back(t->right);
tmp.push_back(t->val);
h--;
}
else
{
t = dq.back();
dq.pop_back();
if(t->right) dq.push_front(t->right);
if(t->left) dq.push_front(t->left);
tmp.push_back(t->val);
h--;
}
}
ans.push_back(tmp);
zz ++;
}
return ans;
}
};
题目来源
