LintCode 71 二叉树的锯齿形层次遍历

题目:solveNQueens

要求:

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:

解题思路:

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<TreeNode*> 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;
}
};