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  • LeetCode: Binary Tree Level Order Traversal

    C++

     1 /**
     2  * Definition for a binary tree node.
     3  * struct TreeNode {
     4  *     int val;
     5  *     TreeNode *left;
     6  *     TreeNode *right;
     7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
     8  * };
     9  */
    10 class Solution {
    11 public:
    12     vector<vector<int>> levelOrder(TreeNode* root) {
    13         vector<vector<int>> ans;
    14         if (root == NULL) return ans;
    15         vector<int> tmp;
    16         queue<TreeNode *> que;
    17         que.push(root);
    18         que.push(NULL);
    19         while (true) {
    20             TreeNode *front = que.front();
    21             que.pop();
    22             if (front == NULL) {
    23                 ans.push_back(tmp);
    24                 if (que.empty() || que.front() == NULL) break;
    25                 tmp.clear();
    26                 que.push(NULL);
    27             }
    28             else {
    29                 if (front->left) que.push(front->left);
    30                 if (front->right) que.push(front->right);
    31                 tmp.push_back(front->val);
    32             }
    33         }
    34         return ans;
    35     }
    36 };
    View Code

    C#代码:

     1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     public int val;
     5  *     public TreeNode left;
     6  *     public TreeNode right;
     7  *     public TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 public class Solution {
    11     public List<List<int>> LevelOrder(TreeNode root) {
    12         List<List<int>> ans = new List<List<int>>();
    13         if (root == null) return ans;
    14         List<int> tmp = new List<int>();
    15         Queue<TreeNode> que = new Queue<TreeNode>();
    16         que.Enqueue(root);
    17         que.Enqueue(null);
    18         while (true)
    19         {
    20             TreeNode peek = que.Peek();
    21             que.Dequeue();
    22             if (peek == null)
    23             {
    24                 ans.Add(tmp);
    25                 if (que.Count == 0 || que.Peek() == null) break;
    26                 tmp = new List<int>();
    27                 que.Enqueue(null);
    28             }
    29             else {
    30                 if (peek.left != null) que.Enqueue(peek.left);
    31                 if (peek.right != null) que.Enqueue(peek.right);
    32                 tmp.Add(peek.val);
    33             }
    34         }
    35         return ans;
    36     }
    37 }
    View Code
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  • 原文地址:https://www.cnblogs.com/yingzhongwen/p/2965451.html
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