LeetCode: Binary Tree Level Order Traversal && Binary Tree Zigzag Level Order Traversal

Title:

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its level order traversal as:

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

思路：使用队列，并且每次记录下每层的数目，这样从队列中取元素也就清楚了每层的元素

/**
 * 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>> levelOrder(TreeNode* root) {
        queue<TreeNode*> Q;
        vector<vector<int> > results;
        if (!root)
            return results;
        vector<int> result;
        Q.push(root);
        int count_pre = 1;
        int count_cur = 0;
        while (!Q.empty()){
            TreeNode* top = Q.front();
            result.push_back(top->val);
            Q.pop();
            count_pre--;
            if (top->left){
                Q.push(top->left);
                count_cur++;
            }
            if (top->right){
                Q.push(top->right);
                count_cur++;
            }
            if (count_pre == 0){
                count_pre = count_cur;
                count_cur = 0;
                results.push_back(result);
                result.clear();
            }
        }
        return results;
    }
};

还可以使用递归。关键是要保存到对应的层数

/**
 * 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>> levelOrder(TreeNode* root) {
        vector<vector<int> > result;
        travel(root,result,1);
        return result;
    }
    void travel(TreeNode* root,vector<vector<int> > &result,int level){
        if (!root)
            return ;
        if (level > result.size())
            result.push_back(vector<int> ());
        result[level-1].push_back(root->val);
        travel(root->left,result,level+1);
        travel(root->right,result,level+1);
    }
};

Title: 

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,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

 

return its zigzag level order traversal as:

[
  [3],
  [20,9],
  [15,7]
]
一样的思路啊！如果不使用递归，则使用一个标记，看这层是否反转，直接使用reverse函数即可。如果使用递归，则也是在相应的层反转。但是因为是在插入的过程中，所以应该是插入每层的vector头。

/**
 * 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> > results;
        if (!root)
            return results;
        queue<TreeNode*> Q;
        Q.push(root);
        int count_pre = 1;
        int count_cur = 0;
        vector<int> result;
        bool flag = false;
        while (!Q.empty()){
            TreeNode* node = Q.front();
            Q.pop();
            result.push_back(node->val);
            count_pre--;
            if (node->left){
                Q.push(node->left);
                count_cur++;
            }
            if (node->right){
                Q.push(node->right);
                count_cur++;
            }
            if (count_pre == 0){
                count_pre = count_cur;
                count_cur = 0;
                if (flag){
                    reverse(result.begin(),result.end());
                    results.push_back(result);
                }else{
                    results.push_back(result);
                }
                result.clear();
                flag = !flag;
            }
        }
        return results;
    }
};

 

/**
 * 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> > results;
        travel(root,1,results,true);
        return results;
    }
    void travel(TreeNode* root, int level, vector<vector<int> > &results, bool left_to_right){
        if (!root)
            return;
        if (level > results.size())
            results.push_back(vector<int>());
        if (left_to_right)
            results[level-1].push_back(root->val);
        else
            results[level-1].insert(results[level-1].begin(),root->val);
        travel(root->left,level+1,results,!left_to_right);
        travel(root->right,level+1,results,!left_to_right);
    }
};