[leetCode]Binary Tree Zigzag Level Order Traversal

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Binary Tree Level Order Traversal
Binary Tree Level Order Traversal II

这两题做下来基本AC没什么难度,题还是连着做顺手。

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
#include <list>
#include <vector>
using namespace std;
struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;     
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
class Solution {
public:
    list<vector<TreeNode>> levelNode;
    vector<vector<int>> allLevelval;
    vector<vector<int>> zigzagLevelOrder(TreeNode *root) {
        vector<TreeNode> curVec;
        vector<int> curIntList;
        if(levelNode.empty() && root != NULL){
            curVec.push_back(*root);
            levelNode.push_back(curVec);
        }
        bool zigzag = true;
        for(list<vector<TreeNode>>::iterator iter = levelNode.begin();iter != levelNode.end();iter++){
            int size = iter->size();
            curVec.clear();
            curIntList.clear();
            for(int i = 0 ; i< size; i++){
                TreeNode curNode = iter->at(i);
                if(curNode.left != NULL) curVec.push_back(*curNode.left);
                if(curNode.right != NULL) curVec.push_back(*curNode.right);

                if(zigzag){
                    curIntList.push_back(curNode.val);
                }else{
                    curIntList.insert(curIntList.begin(),curNode.val);
                }
            }
            allLevelval.push_back(curIntList);

            if(curVec.empty()) break;
            levelNode.push_back(curVec);
            zigzag = !zigzag;
        }
        return allLevelval;
    }
};
艰难的成长
原文地址:https://www.cnblogs.com/marylins/p/3584757.html