103. Binary Tree Zigzag Level Order Traversal (Tree, Queue; BFS)

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,null,null,15,7],

3
/ \
9 20
/ \
15 7

return its zigzag level order traversal as:

[
[3],
[20,9],
[15,7]
]
思路：还是层次遍历，只是增加一个int或bool类型变量标示当前行是否需要reverse。

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<queue<TreeNode*>> q(2);
int curIndex = 0;
int nextIndex = 1;
int levelIndex = 0;
vector<int> retItem;
vector<vector<int>> ret;

if(root) q[curIndex].push(root);
while(!q[curIndex].empty()){
retItem.push_back(q[curIndex].front()->val);
if(q[curIndex].front()->left) q[nextIndex].push(q[curIndex].front()->left);
if(q[curIndex].front()->right) q[nextIndex].push(q[curIndex].front()->right);
q[curIndex].pop();

if(q[curIndex].empty()){ //end of this level
if(levelIndex) reverse(retItem.begin(), retItem.end());
ret.push_back(retItem);
retItem.clear();
curIndex = (curIndex+1) & 0x01;
nextIndex = (nextIndex+1) & 0x01;
levelIndex = (levelIndex+1) & 0x01;
}
}

return ret;
}
};