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题目:Binary Tree Level Order Traversal i ii
i和ii的差别仅在于最后将结果逆序一下就行了,算法上基本相同
个人思路:
1、二叉树的层次遍历,我们一层一层地处理,用一个队列(A队列)将每一层的所有节点按照从左到右的顺序入队
2、待该队列的所有节点都出队,并且用另外一个队列(B队列)接收出队的节点(为下一层节点的处理做准备),则考虑处理下一层节点
3、将B队列的节点逐个出队,考察节点的左孩子和右孩子,若存在则将孩子节点入A队列,如此循环地处理每一层节点,直到A队列为空结束
代码(注释中的代码为i的解):
1 #include <stddef.h> 2 #include <queue> 3 #include <vector> 4 #include <deque> 5 #include <iostream> 6 7 using namespace std; 8 9 struct TreeNode 10 { 11 int val; 12 TreeNode *left; 13 TreeNode *right; 14 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 15 }; 16 17 class Solution 18 { 19 public: 20 vector<vector<int> > levelOrder(TreeNode *root) 21 { 22 /* 23 vector<vector<int> > total_result; 24 vector<int> level_result; 25 queue<TreeNode *> current_level; 26 queue<TreeNode *> current_level_bak; 27 28 if (root == NULL) 29 { 30 return total_result; 31 } 32 33 current_level.push(root); 34 35 while (!current_level.empty()) 36 { 37 TreeNode *current_node = current_level.front(); 38 current_level.pop(); 39 current_level_bak.push(current_node); 40 level_result.push_back(current_node->val); 41 42 if (current_level.empty()) 43 { 44 total_result.push_back(level_result); 45 level_result.clear(); 46 while (!current_level_bak.empty()) 47 { 48 current_node = current_level_bak.front(); 49 current_level_bak.pop(); 50 if (current_node->left != NULL) 51 { 52 current_level.push(current_node->left); 53 } 54 if (current_node->right != NULL) 55 { 56 current_level.push(current_node->right); 57 } 58 } 59 } 60 } 61 62 return total_result; 63 */ 64 65 vector<vector<int> > total_result; 66 deque<vector<int> > temp_result; 67 vector<int> level_result; 68 queue<TreeNode *> current_level; 69 queue<TreeNode *> current_level_bak; 70 71 if (root == NULL) 72 { 73 return total_result; 74 } 75 76 current_level.push(root); 77 78 while (!current_level.empty()) 79 { 80 TreeNode *current_node = current_level.front(); 81 current_level.pop(); 82 current_level_bak.push(current_node); 83 level_result.push_back(current_node->val); 84 85 if (current_level.empty()) 86 { 87 temp_result.push_front(level_result); 88 level_result.clear(); 89 while (!current_level_bak.empty()) 90 { 91 current_node = current_level_bak.front(); 92 current_level_bak.pop(); 93 if (current_node->left != NULL) 94 { 95 current_level.push(current_node->left); 96 } 97 if (current_node->right != NULL) 98 { 99 current_level.push(current_node->right); 100 } 101 } 102 } 103 } 104 105 while (!temp_result.empty()) 106 { 107 total_result.push_back(temp_result.front()); 108 temp_result.pop_front(); 109 } 110 111 return total_result; 112 } 113 }; 114 115 int main() 116 { 117 TreeNode *root = new TreeNode(3); 118 root->left = new TreeNode(9); 119 root->right = new TreeNode(20); 120 root->right->left = new TreeNode(15); 121 root->right->right = new TreeNode(7); 122 123 Solution s; 124 vector<vector<int> > result = s.levelOrder(root); 125 126 for (int i = 0; i < result.size(); ++i) 127 { 128 vector<int> level = result[i]; 129 for (int j = 0; j < level.size(); ++j) 130 { 131 cout << level[j] << ","; 132 } 133 cout << endl; 134 } 135 136 system("pause"); 137 138 return 0; 139 }
网上的思路也大致是这样
leetcode - Binary Tree Level Order Traversal i ii,布布扣,bubuko.com
leetcode - Binary Tree Level Order Traversal i ii
标签:style blog http color os art
原文地址:http://www.cnblogs.com/laihaiteng/p/3797748.html