[LeetCode] Binary Tree Zigzag Level Order Traversal-阿里云开发者社区

[LeetCode] Binary Tree Zigzag Level Order Traversal

2015-07-04 804

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简介： To be honest, I do not know whether this problem is designed to let you use stacks. Anyway, I don't. Here are my codes, both BFS and DFS version.

To be honest, I do not know whether this problem is designed to let you use stacks. Anyway, I don't. Here are my codes, both BFS and DFS version.

 1     // BFS 
 2     vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
 3         vector<vector<int> > levels;
 4         if (!root) return levels;
 5         queue<TreeNode*> toVisit;
 6         toVisit.push(root);
 7         int curLevelNodes = 1;
 8         bool flag = false;
 9         while (!toVisit.empty()) {
10             vector<int> level;
11             for (int i = 0; i < curLevelNodes; i++) {
12                 TreeNode* node = toVisit.front();
13                 toVisit.pop();
14                 level.push_back(node -> val);
15                 if (node -> left) toVisit.push(node -> left);
16                 if (node -> right) toVisit.push(node -> right);
17             }
18             if (flag) {
19                 reverse(level.begin(), level.end());
20                 flag = false;
21             }
22             else flag = true;
23             levels.push_back(level);
24             curLevelNodes = toVisit.size();
25         }
26         return levels;
27     }
28 
29     // DFS
30     void dfs(TreeNode* node, int curLevel, bool& nextLevel, vector<int>& level) {
31         if (!node) return;
32         if (curLevel == 1) {
33             level.push_back(node -> val);
34             if (node -> left || node -> right) nextLevel = true;
35         }
36         else {
37             dfs(node -> left, curLevel - 1, nextLevel, level);
38             dfs(node -> right, curLevel - 1, nextLevel, level);
39         }
40     }
41     vector<vector<int> > zigzagLevelOrder(TreeNode* root) {
42         vector<vector<int> > levels;
43         if (!root) return levels;
44         int curLevel = 1;
45         bool nextLevel = true, flag = false;
46         while (nextLevel) {
47             nextLevel = false;
48             vector<int> level;
49             dfs(root, curLevel++, nextLevel, level);
50             if (flag) {
51                 reverse(level.begin(), level.end());
52                 flag = false;
53             } 
54             else flag = true;
55             levels.push_back(level);
56         }
57         return levels;
58     }

Update

Thanks to the remind of ljdsoft1, I have rewritten the code to traverse the nodes in ziazag-order instead of simply reversing it. The code is as follows, using two stacks.

 1 class Solution { 
 2 public:
 3     vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
 4         vector<vector<int> > levels;
 5         if (!root) return levels;
 6         stack<TreeNode*> curLevel, nextLevel;
 7         curLevel.push(root);
 8         bool reversed = false;
 9         while (!curLevel.empty()) { 
10             vector<int> level;
11             while (!curLevel.empty()) {
12                 TreeNode* node = curLevel.top();
13                 curLevel.pop();
14                 level.push_back(node -> val);
15                 if (reversed) {
16                     if (node -> right) nextLevel.push(node -> right);
17                     if (node -> left) nextLevel.push(node -> left);
18                 }
19                 else {
20                     if (node -> left) nextLevel.push(node -> left);
21                     if (node -> right) nextLevel.push(node -> right);
22                 }
23             }
24             levels.push_back(level);
25             swap(curLevel, nextLevel);
26             reversed = !reversed; 
27         }
28         return levels;
29     }
30 };

[LeetCode] Binary Tree Zigzag Level Order Traversal

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