LeetCode:Binary Tree Level Order Traversal I II

LeetCode:Binary Tree Level Order Traversal

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]
]                                                                     本文地址

队列辅助层序遍历，队列中插入NULL作为层与层之间的间隔，注意处理队列里最后的NULL时，不能再把它入队列以免形成死循环

 /**
  * Definition for binary tree
  * 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) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         vector<vector<int> >res;
         if(root == NULL)return res;
         queue<TreeNode*> myqueue;
         myqueue.push(root);
         myqueue.push(NULL);//NULL是层与层之间间隔标志
         vector<int> level;
         while(myqueue.empty() == false)
         {
             TreeNode *p = myqueue.front();
             myqueue.pop();
             if(p != NULL)
             {
                 level.push_back(p->val);
                 if(p->left)myqueue.push(p->left);
                 if(p->right)myqueue.push(p->right);
             }
             else
             {
                 res.push_back(level);
                 if(myqueue.empty() == false)
                 {
                     level.clear();
                     myqueue.push(NULL);
                 }
             }
         }
         return res;
     }
 };

---

LeetCode:Binary Tree Level Order Traversal II

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

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

    3
   / \
  9  20
    /  \
   15   7

return its bottom-up level order traversal as:

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

和上一题差不多，只是需要把最后遍历结果数组翻转一下

 /**
  * Definition for binary tree
  * struct TreeNode {
  *     int val;
  *     TreeNode *left;
  *     TreeNode *right;
  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
  * };
  */
 class Solution {
 public:
     vector<vector<int> > levelOrderBottom(TreeNode *root) {
         // IMPORTANT: Please reset any member data you declared, as
         // the same Solution instance will be reused for each test case.
         vector<vector<int> >res;
         if(root == NULL)return res;
         queue<TreeNode*> myqueue;
         myqueue.push(root);
         myqueue.push(NULL);//NULL是层与层之间间隔标志
         vector<int> level;
         while(myqueue.empty() == false)
         {
             TreeNode *p = myqueue.front();
             myqueue.pop();
             if(p != NULL)
             {
                 level.push_back(p->val);
                 if(p->left)myqueue.push(p->left);
                 if(p->right)myqueue.push(p->right);
             }
             else
             {
                 res.push_back(level);
                 if(myqueue.empty() == false)
                 {
                     level.clear();
                     myqueue.push(NULL);
                 }
             }
         }
         reverse(res.begin(), res.end());
         return res;
     }
 };

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