Path Sum系列

Path Sum

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 and sum = 22,

              5
             /             4   8
           /   /           11  13  4
         /  \              7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

使用先序遍历，代码简单，如下：

 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 ) return false;   //如果开始root就为空，说明就不存在
14         if( !root->left && !root->right ) return sum == root->val;  //如果是叶节点，那么直接判断sum是否等于叶节点的值
15         if( root->left && hasPathSum(root->left, sum-root->val) ) return true;  //需要确定左子树存在，才继续下搜
16         if( root->right && hasPathSum(root->right, sum-root->val) ) return true;    //同上
17         return false;
18     }
19 };

Path Sum 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 and sum = 22,

              5
             /             4   8
           /   /           11  13  4
         /  \    /         7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
]

同path sum II，依然使用先序遍历，不过加上path变量来记录路径，代码如下：

 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         allpath.clear();
14         if( !root ) return allpath;
15         vector<int> path;
16         preorder(path, root, sum);
17         return allpath;
18     }
19
20     void preorder(vector<int>& path, TreeNode* root, int sum) {
21         if( !root->left && !root->right ) { //如果是叶节点
22             if( sum == root->val ) {    //如果总和刚好是sum，那么放入allpath中
23                 path.push_back(sum);
24                 allpath.push_back(path);
25                 path.pop_back();
26             }
27             return ;
28         }
29         if( root->left ) {  //确保是左子树存在，函数并没有处理空的情况
30             path.push_back(root->val);
31             preorder(path, root->left, sum-root->val);
32             path.pop_back();
33         }
34         if( root->right ) {
35             path.push_back(root->val);
36             preorder(path, root->right, sum-root->val);
37             path.pop_back();
38         }
39     }
40
41 private:
42     vector< vector<int> > allpath;
43 };