LintCode 71 二叉树的锯齿形层次遍历

题目:solveNQueens

---

要求:

给出一棵二叉树，返回其节点值的锯齿形层次遍历（先从左往右，下一层再从右往左，层与层之间交替进行）

样例:

给出一棵二叉树 {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7
返回其锯齿形的层次遍历为：

[
  [3],
  [20,9],
  [15,7]
]

算法要求:

无

解题思路:

在每层判断下是不是该转置。

算法如下:

    int getSize(TreeNode *root) {
        if (root == NULL) {
            return 1;
        } 
        return getSize(root->left) + getSize(root->right);
    }
    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {
        // write your code here
        vector<vector<int> > vec;
        if (root == NULL) {
            return vec;
        }
        int size = getSize(root) - 1;
        queue que;
        que.push(root);
        vector<int> *v = new vector<int>();
        int n = 1;
        int i = 0;
        int j = 0;
        bool flag = true;
        while (!que.empty() && j < size) {
            TreeNode *now = que.front();
            que.pop();
            i++;
            if (now) {
                v->push_back(now->val);
                j++;
            }
            if (i >= n) {
                i = 0;
                n *= 2;
                vector<int> tempVec;
                if (!flag) {
                    vector<int> tempVec2 = *v;
                    int size2 = v->size();
                    for (int i = size2 - 1; i >= 0 ; i--) {
                        tempVec.push_back(tempVec2[i]);
                    }
                }
                flag = !flag;
                if (tempVec.empty()) {
                    tempVec = *v;
                }
                if (!v->empty()) {
                    vec.push_back(tempVec);
                    v = new vector<int>();
                }
            }
            if (now) {
                que.push(now->left);
                que.push(now->right);
            } else{
                que.push(NULL);
                que.push(NULL);
            }
        }
        if (!v->empty()) {
            vector<int> tempVec;
            if (!flag) {
                vector<int> tempVec2 = *v;
                int size2 = v->size();
                for (int i = size2 - 1; i >= 0 ; i--) {
                    tempVec.push_back(tempVec2[i]);
                }
            }
            if (tempVec.empty()) {
                tempVec = *v;
            }
            if (!v->empty()) {
                vec.push_back(tempVec);
                v = new vector<int>();
            }
        }
        return vec;
    }
};