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To be honest, this problem is designed to let you use stacks. However, I don‘t. In fact, you only need to keep a flag and switch it between false and true (initialize to false ). When it is true , reverse the current level before adding it to the result.
Again, BFS and DFS solutions are provided.
BFS:
1 vector<vector<int>> zigzagLevelOrder(TreeNode* root) { 2 vector<vector<int> > levels; 3 if (!root) return levels; 4 queue<TreeNode*> toVisit; 5 toVisit.push(root); 6 int curLevelNodes = 1; 7 bool flag = false; 8 while (!toVisit.empty()) { 9 vector<int> level; 10 for (int i = 0; i < curLevelNodes; i++) { 11 TreeNode* node = toVisit.front(); 12 toVisit.pop(); 13 level.push_back(node -> val); 14 if (node -> left) toVisit.push(node -> left); 15 if (node -> right) toVisit.push(node -> right); 16 } 17 if (flag) { 18 reverse(level.begin(), level.end()); 19 flag = false; 20 } 21 else flag = true; 22 levels.push_back(level); 23 curLevelNodes = toVisit.size(); 24 } 25 return levels; 26 }
DFS:
1 void dfs(TreeNode* node, int curLevel, bool& nextLevel, vector<int>& level) { 2 if (!node) return; 3 if (curLevel == 1) { 4 level.push_back(node -> val); 5 if (node -> left || node -> right) nextLevel = true; 6 } 7 else { 8 dfs(node -> left, curLevel - 1, nextLevel, level); 9 dfs(node -> right, curLevel - 1, nextLevel, level); 10 } 11 } 12 vector<vector<int> > zigzagLevelOrder(TreeNode* root) { 13 vector<vector<int> > levels; 14 if (!root) return levels; 15 int curLevel = 1; 16 bool nextLevel = true, flag = false; 17 while (nextLevel) { 18 nextLevel = false; 19 vector<int> level; 20 dfs(root, curLevel++, nextLevel, level); 21 if (flag) { 22 reverse(level.begin(), level.end()); 23 flag = false; 24 } 25 else flag = true; 26 levels.push_back(level); 27 } 28 return levels; 29 }
[LeetCode] Binary Tree Zigzag Level Order Traversal
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原文地址：http://www.cnblogs.com/jcliBlogger/p/4548038.html
