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 };