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] ]
题目的意思很直白，层序遍历整个树，但是第一层正序输出，第二层反序输出，第三层正序输出，以此类推。做法有两种：一、仍然采用level-travel，只是引入一个标记，判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;
};
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();
}
}
}
};LeetCode :: Binary Tree Zigzag Level Order Traversal [tree, BFS]
原文地址：http://blog.csdn.net/u013195320/article/details/37994615
