二叉树算法总结

Posted 2020-06-29 时间的女儿

刷了几道二叉树的算法题，基本都可以用递归求出来，在可以使用回溯法的题目中，回溯法的时间开销比递归少。

递归调用分为两类：1.在根节点到叶子节点的路径中找出满足条件的值

　　　　　　　　　2.在任意两个节点之间寻找满足条件的路径

根节点到叶子节点的路径选择

leetcode上有类似的题目，

Given a binary tree containing digits from 0-9 only, each root-to-leaf path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the number 123.

Find the total sum of all root-to-leaf numbers.

For example,

    1
   /   2   3

 

The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

思路一：可以使用递归调用，从根节点将需要保存的内容保存起来，判断到达叶子节点则判断该路径是否满足题目要求

 

 

 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     int sumNumbers(TreeNode* root) {
13         vector<int> sumVec;
14         int result;
15         int sum = 0;
16         if(root==NULL){
17             return 0;
18         }
19         sumHelper(root,sumVec,result);
20         for(int i = 0;i<sumVec.size();i++){
21             sum+=sumVec[i];
22         }
23         return sum;
24     }
25     int sumHelper(TreeNode* root,vector<int>& sumVec,int result){
26         if(!root->left&&!root->right){
27             result = result*10+(root->val);
28             sumVec.push_back(result);
29         }
30         if(root->left){
31              int resultl = result;
32              resultl = resultl*10+(root->val);
33              sumHelper(root->left,sumVec,resultl);
34         }
35         if(root->right){
36             result = result*10+(root->val);
37             sumHelper(root->right,sumVec,result);
38         }
39     }
40 };

思路二，回溯法

任意两个节点间的路径选择

Given a binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path does not need to go through the root.

For example:
Given the below binary tree,

       1
      /      2   3

 

Return 6.

使用递归，使用全局变量MaxSum保存历史最大值,以root为起始扩散点的目的是将所有路径遍历完。

注意：节点的值为负数时，不用将其加入路径中。

 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     int maxSum;
12 public:
13     int maxPathSum(TreeNode* root) {
14         maxSum = -99999;
15         if(root==NULL){return 0;}
16         DFS(root);
17         return maxSum;
18     }
19     int DFS(TreeNode* root){
20         if(root==NULL) return 0;
21         int sum = root->val;
22         int l = DFS(root->left);
23         int r = DFS(root->right);
24         if(l>0){
25             sum+=l;
26         }
27         if(r>0){
28             sum+=r;
29         }
30        // int sum = l+r+root->val;
31         maxSum = max(sum,maxSum);
32         return (root->val+max(max(l,0),max(r,0)));
33     }
34 };