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;
    }
};