剑指offer 按之字形顺序打印二叉树

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };
10 */
11 class Solution {
12 public:
13     vector<vector<int> > Print(TreeNode* pRoot) {
14         vector<vector<int> > res;
15         vector<int> temp;
16         int odd = 0; //0表示奇数行，1表示偶数行
17         queue<TreeNode*> q;
18         if (pRoot == NULL)
19             return res;
20         q.push(pRoot);
21         int current = 1, next = 0; //current记录当前打印行的元素数，next表示下一行的元素个数。
22         while (!q.empty()) {
23             temp.clear();
24             for (int i = 0; i < current; i++) {
25                 TreeNode* pNode = q.front();
26                 q.pop();
27                 temp.push_back(pNode->val);
28                 if (pNode->left != NULL) {
29                     q.push(pNode->left);
30                     next++;
31                 }
32                 if (pNode->right != NULL) {
33                     q.push(pNode->right);
34                     next++;
35                 }
36             }
37             //如果是奇数行，就要将保存下来的逆序保存
38             if (odd == 1) {
39                 for (int i = 0; i < (current >> 1); i++) {
40                     swap(temp[i], temp[current - 1 - i]);
41                 }
42             }
43             res.push_back(temp);
44             odd = 1 - odd;
45             current = next;
46             next = 0;
47         }
48         return res;
49     }
50 }；

 1 /*
 2 struct TreeNode {
 3     int val;
 4     struct TreeNode *left;
 5     struct TreeNode *right;
 6     TreeNode(int x) :
 7             val(x), left(NULL), right(NULL) {
 8     }
 9 };
10 */
11 class Solution {
12 public:
13     vector<vector<int> > Print(TreeNode* pRoot) {
14         vector<vector<int> > res;
15         vector<int> temp;
16         stack<TreeNode*> level[2]; 
17         int current = 0, next = 1; 
18         if (pRoot == NULL)
19             return res;
20         level[current].push(pRoot);
21         //如果pRoot == NULL, 那么压入栈之后，栈非空，而取元素的话，又会产生越界？
22         while (!level[current].empty() || !level[next].empty()) {
23             TreeNode* pNode = level[current].top();
24             level[current].pop();
25             temp.push_back(pNode->val);
26             //current == 0表示当前在保存偶数行的数，将下一行的数保存到栈时，先将节点的左孩子放入栈中
27             if (current == 0) {
28                 if (pNode->left != NULL)
29                     level[next].push(pNode->left);
30                 if (pNode->right != NULL)
31                     level[next].push(pNode->right);
32             } else { //当前在保存奇数行的数据，先将节点的右孩子放入栈中
33                 if (pNode->right != NULL)
34                     level[next].push(pNode->right);
35                 if (pNode->left != NULL)
36                     level[next].push(pNode->left);
37             }
38             if (level[current].empty()) {
39                 res.push_back(temp);
40                 temp.clear();
41                 current = 1 - current;
42                 next = 1 - next;
43             }
44         }
45         return res;
46     }
47 };