LeetCode(102)Binary Tree Level Order Traversal

题目如下:

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]
]
分析如下:
本题考察二叉树的层序遍历,之前也 写过类似的主题。本题我尝试着不在IDE中写,而是直接在Leetcode网页提供的编辑器上写。最后自认为非常小心翼翼地检查好了,在IDE中跑了跑,发现挂了,走入了死循环。于是调试之,找bug之。终于提交。整体花了60分钟左右。远远低于面试要求平均每道题25分钟搞定并且bug-free的要求,哎。哎。
//20ms过大集合
/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
#include <queue>;
using namespace std;
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        vector<vector<int>> res_vec_vec;
        if(root==NULL)
            return res_vec_vec;
        queue<queue<TreeNode*>> que_que;
        queue<TreeNode*> que;
        que.push(root);
        que_que.push(que);
        while(!que_que.empty()){
            queue<TreeNode*> tmp_src_queue=que_que.front();
            queue<TreeNode*> tmp_out_queue;
            vector<int> tmp_vec;
            while(!tmp_src_queue.empty()) {
                tmp_vec.push_back(tmp_src_queue.front()->val);
                TreeNode* tmp_node=tmp_src_queue.front();
                if(tmp_node->left!=NULL)
                    tmp_out_queue.push(tmp_node->left);
                if(tmp_node->right!=NULL)
                    tmp_out_queue.push(tmp_node->right);
                tmp_src_queue.pop();
            }
            res_vec_vec.push_back(tmp_vec);
            if(!tmp_out_queue.empty())//如果没有这个条件判断,程序就死循环了
                que_que.push(tmp_out_queue);
            que_que.pop();
        }
        return res_vec_vec;
    }

};

小结:
(1) bug-free很重要,现在就要开始练习,要不然面试必挂。
(2) bug-free的实际意义是,在真正的开发中,省去找bug的时间。与其你写个漏洞百出的code然后等着一点一点地被修正,花去大量时间。不如从一开始就像清楚,写清楚。
(3) 本题的一个重要bug是下面的这句话
            if(!tmp_out_queue.empty())//如果没有这个条件判断,程序就死循环了
                que_que.push(tmp_out_queue);
之前没有写if()条件就直接push了,导致死循环。这个bug的实质在于,对于循环的边界条件的判断还是做得不够好,两层循环一出来,立刻暴露缺陷。

update : 2015-03-24
受到 Subset这道题目的启发,用size记录下当前level的节点个数,这样就可以简化很多代码。
//11ms 
class Solution {
public:
    vector<vector<int> > levelOrder(TreeNode *root) {
        queue<TreeNode*> q;
        if (root != NULL) {
            q.push(root);
        }
        vector<int> inner;
        vector<vector<int> > outer;
        TreeNode* node;
        while(!q.empty()) {
            int size = q.size();
            inner.clear();
            while (size-- > 0) {
                node = q.front();
                q.pop();
                inner.push_back(node->val);
                if (node->left != NULL)
                    q.push(node->left);
                if (node->right != NULL)
                    q.push(node->right);
            }
            outer.push_back(inner);
        }
        return outer;
    }
};



1. Two Sum 2. Add Two Numbers 3. Longest Substring Without Repeating Characters 4. Median of Two Sorted Arrays 5. Longest Palindromic Substring 6. ZigZag Conversion 7. Reverse Integer 8. String to Integer (atoi) 9. Palindrome Number 10. Regular Expression Matching 11. Container With Most Water 12. Integer to Roman 13. Roman to Integer 14. Longest Common Prefix 15. 3Sum 16. 3Sum Closest 17. Letter Combinations of a Phone Number 18. 4Sum 19. Remove Nth Node From End of List 20. Valid Parentheses 21. Merge Two Sorted Lists 22. Generate Parentheses 23. Swap Nodes in Pairs 24. Reverse Nodes in k-Group 25. Remove Duplicates from Sorted Array 26. Remove Element 27. Implement strStr() 28. Divide Two Integers 29. Substring with Concatenation of All Words 30. Next Permutation 31. Longest Valid Parentheses 32. Search in Rotated Sorted Array 33. Search for a Range 34. Find First and Last Position of Element in Sorted Array 35. Valid Sudoku 36. Sudoku Solver 37. Count and Say 38. Combination Sum 39. Combination Sum II 40. First Missing Positive 41. Trapping Rain Water 42. Jump Game 43. Merge Intervals 44. Insert Interval 45. Unique Paths 46. Minimum Path Sum 47. Climbing Stairs 48. Permutations 49. Permutations II 50. Rotate Image 51. Group Anagrams 52. Pow(x, n) 53. Maximum Subarray 54. Spiral Matrix 55. Jump Game II 56. Merge k Sorted Lists 57. Insertion Sort List 58. Sort List 59. Largest Rectangle in Histogram 60. Valid Number 61. Word Search 62. Minimum Window Substring 63. Unique Binary Search Trees 64. Unique Binary Search Trees II 65. Interleaving String 66. Maximum Product Subarray 67. Binary Tree Inorder Traversal 68. Binary Tree Preorder Traversal 69. Binary Tree Postorder Traversal 70. Flatten Binary Tree to Linked List 71. Construct Binary Tree from Preorder and Inorder Traversal 72. Construct Binary Tree from Inorder and Postorder Traversal 73. Binary Tree Level Order Traversal 74. Binary Tree Zigzag Level Order Traversal 75. Convert Sorted Array to Binary Search Tree 76. Convert Sorted List to Binary Search Tree 77. Recover Binary Search Tree 78. Sum Root to Leaf Numbers 79. Path Sum 80. Path Sum II 81. Binary Tree Maximum Path Sum 82. Populating Next Right Pointers in Each Node 83. Populating Next Right Pointers in Each Node II 84. Reverse Linked List 85. Reverse Linked List II 86. Partition List 87. Rotate List 88. Remove Duplicates from Sorted List 89. Remove Duplicates from Sorted List II 90. Intersection of Two Linked Lists 91. Linked List Cycle 92. Linked List Cycle II 93. Reorder List 94. Binary Tree Upside Down 95. Binary Tree Right Side View 96. Palindrome Linked List 97. Convert Binary Search Tree to Sorted Doubly Linked List 98. Lowest Common Ancestor of a Binary Tree 99. Lowest Common Ancestor of a Binary Search Tree 100. Binary Tree Level Order Traversal II
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值