学步园

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]
]

思路：层序遍历的改进。

/**
 * Definition for binary tree
 * 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) {
        deque<TreeNode*> Q1,Q2;
        vector<vector<int> > res;
        res.clear();
        if (root == NULL) {
            return res;
        }
        Q1.push_back(root);
        vector<int> perRes;
        perRes.push_back(Q1.back()->val);
        res.push_back(perRes);
        int i = 0;
        while(!Q1.empty()) {
            Q2.clear();
            while(!Q1.empty()) {
                TreeNode* pNode = Q1.front();
                Q1.pop_front();
                if (pNode->left != NULL) {
                    Q2.push_back(pNode->left);
                }
                if (pNode->right != NULL) {
                    Q2.push_back(pNode->right);
                }
            }
            perRes.clear();
            i++;
            Q1 = Q2;
            if (i&1) {
                while(!Q2.empty()) {
                    perRes.push_back(Q2.back()->val);
                    Q2.pop_back();
                }
            }
            else {
                while(!Q2.empty()) {
                    perRes.push_back(Q2.front()->val);
                    Q2.pop_front();
                }
            }
            if (!perRes.empty()) {
                res.push_back(perRes);
            }
        }
        return res;
    }
};