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O(n)O(n)O(n)

/*** 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;}
};

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