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二叉树与双向链表问题

2024-09-16 10:10:12   218人阅读

技术分享

做完这道题，我深刻认识到通过边画图边写代码揣摩分析才能真正弄懂这道题，而且如果不是独立思考AC看了题解恐怕就很难弄懂这道题。

这里整理一下题解：本题思路有两种：递归和非递归

Java版AC代码：

数据结构定义：

 1 /**
 2 public class TreeNode {
 3     int val = 0;
 4     TreeNode left = null;
 5     TreeNode right = null;
 6 
 7     public TreeNode(int val) {
 8         this.val = val;
 9 
10     }
11 }
12 */

递归版：

 1 public class Solution {
 2     public TreeNode Convert(TreeNode pRootOfTree) {
 3         if(pRootOfTree == null)
 4             return null;
 5         TreeNode res = pRootOfTree;
 6         if(pRootOfTree.right != null) {
 7             TreeNode tRight = Convert(pRootOfTree.right);
 8             pRootOfTree.right = tRight;
 9             tRight.left = pRootOfTree;
10         }
11              
12         if(pRootOfTree.left != null) {
13             TreeNode tLeft = Convert(pRootOfTree.left);
14             TreeNode tmp = tLeft;
15             while(tmp.right != null) {
16                 tmp = tmp.right;
17             }
18             tmp.right = pRootOfTree;
19             pRootOfTree.left = tmp;
20             res = tLeft;
21         }
22         return res;
23 }

非递归版：

技术分享

Java中有关Stack的API

技术分享

 1 import java.util.Stack;
 2 public class Solution {//画图容易理解
 3     public TreeNode Convert(TreeNode pRootOfTree) {
 4         if(pRootOfTree == null)
 5             return null;
 6         Stack<TreeNode> st = new Stack<TreeNode>();
 7         TreeNode tnp, tmp = pRootOfTree;
 8         TreeNode res = tmp;
 9         st.push(tmp);
10         while(tmp.left != null) {
11             tmp = tmp.left;
12             st.push(tmp);
13         }
14         
15         if(!st.empty()) 
16             res = st.peek();
17         
18         while(!st.empty()) {
19             tmp = st.pop();
20             tnp = tmp.right;
21             if(tnp != null) {
22                 st.push(tnp);
23                 while(tnp.left != null) {
24                     tnp = tnp.left;
25                     st.push(tnp);
26                 }
27                 tnp.left = tmp;
28                 tmp.right = tnp;
29             } else {
30                 if(!st.empty()) {//peek()操作前一定要进行判空操作，否则会产生EmptyStack异常
31                     tmp.right = st.peek();
32                     tmp.right.left = tmp;
33                 }
34             }
35         }
36         return res;
37     }
38 }

C++版AC代码：

数据结构：

1 /*
2 struct TreeNode {
3     int val;
4     struct TreeNode *left;
5     struct TreeNode *right;
6     TreeNode(int x) :
7             val(x), left(NULL), right(NULL) {
8     }
9 };*/

递归版：

 1 class Solution {
 2 public:
 3     TreeNode* Convert(TreeNode* pRootOfTree)
 4     {
 5         if(pRootOfTree == NULL)
 6             return NULL;
 7         TreeNode* res = pRootOfTree;
 8         TreeNode* tmp = res;
 9         if(tmp->right != NULL) {
10             TreeNode* tRight = Convert(tmp->right);
11             pRootOfTree->right = tRight;
12             tRight->left = pRootOfTree;
13         }
14         
15         if(tmp->left != NULL) {
16             TreeNode* tLeft = Convert(tmp->left);
17             res = tLeft;
18             while(tLeft->right != NULL) {
19                 tLeft = tLeft->right;
20             }
21             tLeft->right = pRootOfTree;
22             pRootOfTree->left = tLeft;
23         }
24         
25         return res;
26     }
27 };

非递归版：

C++关于stack(首字母小写）的API

  1. bool empty();//判空
  2. void pop();//出栈
  3. void push(const Type &val);//入栈
  4. size_type size();//得到栈的大小
  5. Type &top();//取得栈顶元素引用但不弹出
 1 #include <bits/stdc++.h>
 2 class Solution {
 3 public:
 4     TreeNode* Convert(TreeNode* pRootOfTree)
 5     {
 6         if(pRootOfTree == NULL)
 7             return NULL;
 8         stack<TreeNode*> st;
 9         st.push(pRootOfTree);
10         TreeNode* tmp = pRootOfTree;
11         TreeNode* tnp, *res = tmp;
12         while(tmp->left != NULL) {
13             tmp = tmp->left;
14             st.push(tmp);
15         }
16         
17         if(!st.empty()) {
18             res = st.top();
19         }
20         
21         while(!st.empty()) {
22             tmp = st.top();
23             st.pop();
24             tnp = tmp->right;
25             if(tnp != NULL) {
26                 st.push(tnp);
27                 while(tnp->left != NULL) {
28                     tnp = tnp->left;
29                     st.push(tnp);
30                 }
31                 tnp->left = tmp;
32                 tmp->right = tnp;
33             } else {
34                 if(!st.empty()) {//一定注意要判空
35                     tmp->right = st.top();
36                     tmp->right->left = tmp;
37                 }
38             }
39         }
40         return res;
41     }
42 };

 

二叉树与双向链表问题


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