LeetCode(Binary Tree Level Order Traversal, 2,Zigzag)二叉树的层次遍历

1，题目要求：

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]
]

2.

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],
]

3.

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]
]

代码：

class Solution {
public:
vector<vector<int> > ans;
vector<vector<int> > levelOrder(TreeNode *root) {

if (NULL == root) {
return ans;
}
vector<TreeNode*> q;
int next_level_cnt = 1;//记录下一层节点个数
int level_start = 0;//根节点索引从o开始
int level_idx = 1;
q.push_back(root);
while (level_start < q.size())
{
vector<int> level;
int cur_level_cnt = next_level_cnt;
int level_end = level_start + cur_level_cnt;//计算当前层的结束位置
next_level_cnt = 0;
for (size_t i = level_start; i < level_end; ++i)
{
TreeNode* node = q[i];
level.push_back(node->val);

if(node->left != NULL)
{
q.push_back(node->left);
++next_level_cnt;
}
if(node->right != NULL)
{
q.push_back(node->right);
++next_level_cnt;
}
}
ans.push_back(level);
level_start += cur_level_cnt;//下层的开始位置
}
return ans;
}

vector<vector<int> > zigzagLevelOrder(TreeNode *root) {//用两个栈逐层

if (NULL ==  root) {
return ans;
}
stack<TreeNode*> even_st;
stack<TreeNode*> odd_st;
int level_idx = 1;
int next_level_cnt = 1;
int level_start = 0;
odd_st.push(root);
while (!odd_st.empty() || !even_st.empty()) {
vector<int> level;
//            cout << "odd_st size : " << odd_st.size() << "  even st size : " << even_st.size() << endl;
if((level_idx & 1) == 1)//奇数层,将下一层的节点压入偶数栈
{
while (!odd_st.empty()) {
TreeNode* node = odd_st.top();
odd_st.pop();
level.push_back(node->val);
if(node->left != NULL)
even_st.push(node->left);
if(node->right != NULL)
even_st.push(node->right);
}
}
else//如果是偶数层，将下一层节点压入奇数栈，并且先压入又孩子，再压入左孩子
{
while (!even_st.empty()) {
TreeNode* node = even_st.top();
even_st.pop();
level.push_back(node->val);
if(node->right != NULL)
odd_st.push(node->right);
if(node->left != NULL)
odd_st.push(node->left);

}
}
++level_idx;
ans.push_back(level);
}
return ans;
}
};