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]
]

题目的意思非常直白。层序遍历整个树,可是第一层正序输出。第二层反序输出,第三层正序输出,以此类推。做法有两种:一、仍然採用level-travel,仅仅是引入一个标记,推断是否反转得到的数列; 二、考虑到stack的特点,利用stack FILO的特点来直接输出。两种方法都贴出来

利用stack的:

class Solution {
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
bool isRe = false;
vector<int> a;
stack<TreeNode *> s1, s2; if (root == NULL)
return ret;
s1.push(root);
while (!s1.empty()){
TreeNode *tmp = s1.top();
s1.pop();
a.push_back(tmp->val); if (isRe){
if (tmp->right)
s2.push(tmp->right);
if (tmp->left)
s2.push(tmp->left);
}
else{
if (tmp->left)
s2.push(tmp->left);
if (tmp->right)
s2.push(tmp->right);
}
if (s1.empty()){
ret.push_back(a);
isRe = !isRe;
swap(s1, s2);
a.clear();
}
}
return ret;
}
private:
vector<vector<int>> ret;
};

利用queue的,这里因为引入了swap,所以能够复用同一个代码流程,代码会短一些;

class Solution {
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
vector<vector<int>> ret;
queue<TreeNode *> current, next; //利用两个队列的交替来区分每一层
bool isRe = false;
vector<int> v; if (root == NULL)
return ret;
current.push(root);
while (!current.empty()){
TreeNode *tmp = current.front();
current.pop();
v.push_back(tmp->val); if (tmp->left)
next.push(tmp->left);
if (tmp->right)
next.push(tmp->right); if(current.empty()){
if (isRe){
reverse(v.begin(), v.end());
}
ret.push_back(v);
swap(current,next);
isRe = !isRe;
v.clear();
}
} }
};

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