leetcode_c++：栈：Binary Tree Zigzag Level Order Traversal(103)

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]
]

//求层次遍历，但是奇数层要是反着的输出

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http://www.acmerblog.com/leetcode-solution-binary-tree-zigzag-level-order-traversal-6239.html

//广度优先遍历，用一个bool记录是从左到右还是从右到左，每一层结束就翻转一下。

// 迭代版，时间复杂度O(n)，空间复杂度O(n)

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#include <iostream>#include <vector>#include<cmath>#include <algorithm>#include <stack>#include <string>#include <sstream>#include <queue>using namespace std;struct TreeNode{    int val;    TreeNode *left;    TreeNode *right;    TreeNode(int x): val(x),left(NULL),right(NULL){}};//LeetCode, Binary Tree Zigzag Level Order Traversal//广度优先遍历，用一个bool记录是从左到右还是从右到左，每一层结束就翻转一下。// 迭代版，时间复杂度O(n)，空间复杂度O(n)class Solution {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        vector<vector<int> > result;        if (nullptr == root) return result;        queue<TreeNode*> q;        bool left_to_right = true;  //left to right        vector<int> level;  // one level's elements        q.push(root);        q.push(nullptr);  // level separator        while (!q.empty()) {            TreeNode *cur = q.front();            q.pop();            if (cur) {                level.push_back(cur->val);                if (cur->left) q.push(cur->left);                if (cur->right) q.push(cur->right);            } else {                if (left_to_right) {                    result.push_back(level);                } else {                    reverse(level.begin(), level.end());                    result.push_back(level);                }                level.clear();                left_to_right = !left_to_right;                if (q.size() > 0) q.push(nullptr);            }        }        return result;    }};TreeNode* CreateTree(int a[],int n){    if(n<=0)  return NULL;    TreeNode** tree=new TreeNode*[n];    for(int i=0;i<n;i++){        if(a[i]==0){            tree[i]=NULL;            continue;        }        tree[i]=new TreeNode(a[i]);    }    int pos=1;    for(int i=0;i<n && pos<n;i++){        if(tree[i]){            tree[i]->left=tree[pos++];            if(pos<n){                tree[i]->right=tree[pos++];            }        }    }    return tree[0];}void printTree_level_order(TreeNode* root){    queue<TreeNode*> q;    q.push(root);    while(q.size()>0){        TreeNode* n=q.front();        q.pop();        if(n==NULL){            cout<<"# ";            continue;        }        cout<<n->val<<" ";        q.push(n->left);        q.push(n->right);    }    cout<<endl;}int printMatrix(vector<vector<int> > &vv){    for(int i=0;i<vv.size();i++){        cout<<"[";        for(int j=0;j<vv[i].size();j++)            cout<<" "<<vv[i][j];        cout<<"]"<<endl;    }}int main(int argc,char** argv){    Solution st;    TreeNode* p;    vector<vector<int> > vv;    int a[]={3,9,20,0,0,15,7};    p=CreateTree(a,sizeof(a)/sizeof(int));    printTree_level_order(p);    vv=st.zigzagLevelOrder(p);    printMatrix(vv);    cout<<endl;}

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递归版本

// LeetCode, Binary Tree Zigzag Level Order Traversal// 递归版，时间复杂度O(n)，空间复杂度O(n)class Solution {public:    vector<vector<int> > zigzagLevelOrder(TreeNode *root) {        vector<vector<int>> result;        traverse(root, 1, result, true);        return result;    }    void traverse(TreeNode *root, size_t level, vector<vector<int>> &result,            bool left_to_right) {        if (!root) return;        if (level > result.size())            result.push_back(vector<int>());        if (left_to_right)            result[level-1].push_back(root->val);        else            result[level-1].insert(result[level-1].begin(), root->val);        traverse(root->left, level+1, result, !left_to_right);        traverse(root->right, level+1, result, !left_to_right);    }};