标签：pre treenode val ack path div binary init 递归
I、Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree andsum = 22,
5
/ 4 8
/ / 11 13 4
/ \ 7 2 1
return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.
PS：要想好判断条件，递归的时候考虑到必须要到子节点。
1 /** 2 * Definition for binary tree 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 bool hasPathSum(TreeNode *root, int sum) { 13 if(root==NULL) return false; 14 return dfs(root,sum); 15 } 16 bool dfs(TreeNode *root, int sum){ 17 if(root==NULL) return false; 18 if(root->left==NULL&&root->right==NULL&&sum==root->val) return true; 19 return dfs(root->left,sum-root->val)||dfs(root->right,sum-root->val); 20 } 21 };
II、
Given a binary tree and a sum, find all root-to-leaf paths where each path‘s sum equals the given sum.
For example:
Given the below binary tree andsum = 22,
5
/ 4 8
/ / 11 13 4
/ \ / 7 2 5 1
return
[ [5,4,11,2], [5,8,4,5] ]
1 /** 2 * Definition for binary tree 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 vector<vector<int> > pathSum(TreeNode *root, int sum) { 13 dfs(root,sum); 14 return res; 15 } 16 void dfs(TreeNode *root, int sum){ 17 if(root==NULL) return; 18 v.push_back(root->val); 19 if(root->left==NULL&&root->right==NULL&&root->val==sum){ 20 res.push_back(v); 21 }else{ 22 dfs(root->left,sum-root->val); 23 dfs(root->right,sum-root->val); 24 } 25 v.pop_back(); 26 } 27 vector<int> v; 28 vector<vector<int>> res; 29 };
标签：pre treenode val ack path div binary init 递归
原文地址：http://www.cnblogs.com/zl1991/p/7008194.html
