Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/
9 20
/
15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
思路
宽搜的变形，在宽搜的基础上加一个栈，用于在偶数行反向输出。
1 vector<vector<int> > zigzagLevelOrder(TreeNode *root) { 2 // Note: The Solution object is instantiated only once and is reused by each test case. 3 vector<vector<int> > result; 4 if(root == NULL) 5 return result; 6 vector<int> tmp; 7 int type = 1; 8 TreeNode * tPoint; 9 stack<int> aStack; 10 queue<TreeNode*> aQueue1, aQueue2; 11 aQueue1.push(root); 12 while(!aQueue1.empty() || !aQueue2.empty()){ 13 tmp.clear(); 14 if(type){ 15 while(!aQueue1.empty()){ 16 tPoint = aQueue1.front(); 17 aQueue1.pop(); 18 tmp.push_back(tPoint->val); 19 if(tPoint->left != NULL) 20 aQueue2.push(tPoint->left); 21 if(tPoint->right != NULL) 22 aQueue2.push(tPoint->right); 23 } 24 result.push_back(tmp); 25 type = (type+1)%2; 26 } 27 else{ 28 while(!aQueue2.empty()){ 29 tPoint = aQueue2.front(); 30 aQueue2.pop(); 31 aStack.push(tPoint->val); 32 if(tPoint->left != NULL) 33 aQueue1.push(tPoint->left); 34 if(tPoint->right != NULL) 35 aQueue1.push(tPoint->right); 36 } 37 while(!aStack.empty()){ 38 tmp.push_back(aStack.top()); 39 aStack.pop(); 40 } 41 result.push_back(tmp); 42 type = (type+1)%2; 43 } 44 } 45 return result; 46 }
第二遍，交替使用两个栈
1 vector<vector<int> > zigzagLevelOrder(TreeNode *root) { 2 // Note: The Solution object is instantiated only once and is reused by each test case. 3 vector<vector<int> > result; 4 vector<int> tmp; 5 if(!root) 6 return result; 7 stack<TreeNode *> stacks[2]; 8 bool flag = false; 9 stacks[0].push(root); 10 while(!stacks[flag].empty() || !stacks[!flag].empty()){ 11 while(!stacks[flag].empty()){ 12 TreeNode *node = stacks[flag].top(); 13 stacks[flag].pop(); 14 tmp.push_back(node->val); 15 if(!flag){ 16 if(node->left) 17 stacks[!flag].push(node->left); 18 if(node->right) 19 stacks[!flag].push(node->right); 20 } 21 else{ 22 if(node->right) 23 stacks[!flag].push(node->right); 24 if(node->left) 25 stacks[!flag].push(node->left); 26 } 27 } 28 result.push_back(tmp); 29 tmp.clear(); 30 flag = !flag; 31 } 32 return result; 33 }