标签：des style class blog code http
题目：Flatten Binary Tree to Linked List
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
1
/ 2 5
/ \ 3 4 6
The flattened tree should look like:
1
2
3
4
5
6
If you notice carefully in the flattened tree, each node‘s right child points to the next node of a pre-order traversal.
个人思路：
1、创建一个队列，把节点按照先序遍历的顺序入队
2、把节点挨个出队，前一个出队节点是后一个出队节点的父节点，后一个出队节点是前一个出队节点的右孩子节点，且每个非叶节点的左孩子为空
代码：
1 #include <stddef.h> 2 #include <iostream> 3 #include <queue> 4 5 using namespace std; 6 7 struct TreeNode 8 { 9 int val; 10 TreeNode *left; 11 TreeNode *right; 12 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 13 }; 14 15 class Solution 16 { 17 public: 18 void flatten(TreeNode *root) 19 { 20 if (!root) 21 { 22 return; 23 } 24 25 queue<TreeNode *> preNodes; 26 preorder(root, preNodes); 27 root = preNodes.front(); 28 preNodes.pop(); 29 TreeNode * currentNode = root; 30 while (!preNodes.empty()) 31 { 32 TreeNode *tmp = preNodes.front(); 33 preNodes.pop(); 34 currentNode->right = tmp; 35 currentNode->left = NULL; 36 currentNode = currentNode->right; 37 } 38 } 39 private: 40 void preorder(TreeNode *root, queue<TreeNode *> &preNodes) 41 { 42 if (!root) 43 { 44 return; 45 } 46 preNodes.push(root); 47 preorder(root->left, preNodes); 48 preorder(root->right, preNodes); 49 } 50 };
在网上看到一个比较普遍的递归算法，链接：http://blog.unieagle.net/2012/12/16/leetcode-problemflatten-binary-tree-to-linked-list/，里面有比较详细的思路解析，可以参考，大致思路是这样：要扁平化一棵树，则扁平化根节点的左子树和右子树，然后将左子树放到根节点的右孩子位置，右子树放到左子树的叶节点的右孩子位置
代码：
1 #include <stddef.h> 2 #include <iostream> 3 #include <queue> 4 5 using namespace std; 6 7 struct TreeNode 8 { 9 int val; 10 TreeNode *left; 11 TreeNode *right; 12 TreeNode(int x) : val(x), left(NULL), right(NULL) {} 13 }; 14 15 class Solution 16 { 17 public: 18 void flatten(TreeNode *root) 19 { 20 /* 21 if (!root) 22 { 23 return; 24 } 25 26 queue<TreeNode *> preNodes; 27 preorder(root, preNodes); 28 root = preNodes.front(); 29 preNodes.pop(); 30 TreeNode * currentNode = root; 31 while (!preNodes.empty()) 32 { 33 TreeNode *tmp = preNodes.front(); 34 preNodes.pop(); 35 currentNode->right = tmp; 36 currentNode->left = NULL; 37 currentNode = currentNode->right; 38 } 39 */ 40 41 if (!root) 42 { 43 return; 44 } 45 if (root->left) 46 { 47 flatten(root->left); 48 } 49 if (root->right) 50 { 51 flatten(root->right); 52 } 53 if (!root->left) 54 { 55 return; 56 } 57 58 TreeNode *left_leaf = root->left; 59 while (left_leaf->right) 60 { 61 left_leaf = left_leaf->right; 62 } 63 left_leaf->right = root->right; 64 root->right = root->left; 65 root->left = NULL; 66 } 67 private: 68 void preorder(TreeNode *root, queue<TreeNode *> &preNodes) 69 { 70 if (!root) 71 { 72 return; 73 } 74 preNodes.push(root); 75 preorder(root->left, preNodes); 76 preorder(root->right, preNodes); 77 } 78 };
链接中还有一个非递归的思路，大家也可以看看
leetcode - Flatten Binary Tree to Linked List,布布扣,bubuko.com
leetcode - Flatten Binary Tree to Linked List
标签：des style class blog code http
原文地址：http://www.cnblogs.com/laihaiteng/p/3799222.html
