标签：leetcode 二叉树 bfs 层序遍历 queue
problem：
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.
thinking：
（1）层序遍历采用BFS广度优先搜索，借用队列queue
（2）要实现Z字形 层序遍历，要采用一个BOOL 值标记先访问左还是右孩子的顺序，还要借助两个queue，
其中一个用于输出VAL值，另外一个要借助stack实现翻转，保存访问下一层的孩子节点
code：
class Solution {
private:
vector<vector<int> > ret;
public:
vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
ret.clear();
if(root==NULL)
return ret;
queue<TreeNode *> tmp_queue;
tmp_queue.push(root);
level_order(tmp_queue,false);
return ret;
}
protected:
void level_order(queue<TreeNode *> &queue1,bool flag)
{
if(queue1.empty())
return;
vector<int> array;
queue<TreeNode *> queue0=queue1;
queue<TreeNode *> queue2=queue_reverse(queue1);
queue<TreeNode *> queue3;
while(!queue0.empty()) //打印结点
{
TreeNode *tmp=queue0.front();
queue0.pop();
array.push_back(tmp->val);
}
while(!queue2.empty()) //访问孩子节点
{
TreeNode *tmp=queue2.front();
queue2.pop();
if(flag) //flag规定先访问左孩子还是右孩子
{
if(tmp->left!=NULL)
queue3.push(tmp->left);
if(tmp->right!=NULL)
queue3.push(tmp->right);
}
else
{
if(tmp->right!=NULL)
queue3.push(tmp->right);
if(tmp->left!=NULL)
queue3.push(tmp->left);
}
}
flag=!flag;
ret.push_back(array);
level_order(queue3,flag); //递归访问下一层
}
queue<TreeNode *> queue_reverse(queue<TreeNode *> &_queue) //借助stack实现queue翻转
{
stack<TreeNode *> _stack;
queue<TreeNode *> ret_queue;
while(!_queue.empty())
{
TreeNode *tmp=_queue.front();
_stack.push(tmp);
_queue.pop();
}
while(!_stack.empty())
{
TreeNode *tmp=_stack.top();
ret_queue.push(tmp);
_stack.pop();
}
return ret_queue;
}
};leetcode || 103、Binary Tree Zigzag Level Order Traversal
标签：leetcode 二叉树 bfs 层序遍历 queue
原文地址：http://blog.csdn.net/hustyangju/article/details/45165817
