[leetcode] 103. Binary Tree Zigzag Level Order Traversal

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]

解法一：

加一個flag變量記錄是否reverse。

/**
 * 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;
        queue<TreeNode*> q;
        q.push(root);
        
        bool reverse_flag = false;
        while(!q.empty()){
            int sz = q.size();
            vector<int> level;
            for(int i=0; i<sz; i++){
                TreeNode* tmp = q.front();
                q.pop();
                level.push_back(tmp->val);
                if(tmp->left) q.push(tmp->left);
                if(tmp->right) q.push(tmp->right);
            }
            if(reverse_flag){
                reverse(level.begin(),level.end());
            }
            res.push_back(level);
            reverse_flag = !reverse_flag;
        }
        return res;
        
    }
};

解法二：

/**
 * 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;
        stack<TreeNode*> s1;  // left->right;
        stack<TreeNode*> s2; // right->left;
        s1.push(root);
        
        while(!s1.empty() || !s2.empty()){
            vector<int> out;
            while(!s1.empty()){
                TreeNode* n = s1.top();
                s1.pop();
                out.push_back(n->val);
                if(n->left) s2.push(n->left);
                if(n->right) s2.push(n->right);
            }
            if (!out.empty()) res.push_back(out);
            out.clear();
            
            while(!s2.empty()){
                TreeNode* n = s2.top();
                s2.pop();
                out.push_back(n->val);
                if(n->right) s1.push(n->right);
                if(n->left) s1.push(n->left);
            }            
            if (!out.empty()) res.push_back(out);
        }

        return res;
        
    }
};