LeetCode 103. Binary Tree Zigzag Level Order Traversal
Solution1:基于层次遍历的微改
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int> > res;
if (!root) return res;
vector<int> temp;
queue<TreeNode* > my_que; //队列用来暂时存储二叉树节点
my_que.push(root);
int cur = 1, next = 0, symbol = 0;
while (!my_que.empty()) {
temp.push_back(my_que.front()->val);
cur--;
if (my_que.front()->left) {
my_que.push(my_que.front()->left);
next++;
}
if (my_que.front()->right) {
my_que.push(my_que.front()->right);
next++;
}
my_que.pop();
if (cur == 0) {
if (symbol % 2 == 1) {
reverse(temp.begin(), temp.end());
res.push_back(temp);
} else res.push_back(temp);
temp.clear();
cur = next;
next = 0;
symbol++;
}
}
return res;
}
};
Solution2:用两个栈来保存数据,时间复杂度为
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> zigzagLevelOrder(TreeNode* root) {
vector<vector<int> > res;
if(root == NULL)
return res;
vector<int> temp;
stack<struct TreeNode* > stack_node[2];//栈数组,数组的每个元素均是栈,栈的数据类型是TreeNode*
int cur = 0, next = 1;
stack_node[cur].push(root);
while(!stack_node[0].empty() || !stack_node[1].empty()) {
temp.push_back(stack_node[cur].top()->val);
if(cur == 0) {
if(stack_node[cur].top()->left != NULL)
stack_node[next].push(stack_node[cur].top()->left);
if(stack_node[cur].top()->right != NULL)
stack_node[next].push(stack_node[cur].top()->right);
} else {
if(stack_node[cur].top()->right != NULL)
stack_node[next].push(stack_node[cur].top()->right);
if(stack_node[cur].top()->left != NULL)
stack_node[next].push(stack_node[cur].top()->left);
}
stack_node[cur].pop();
if(stack_node[cur].empty()) {
res.push_back(temp);
temp.clear();
cur = 1 - cur;
next = 1 - next;
}
}
return res;
}
};