Path Sum & Path Sum II | LeetCode DFS

Description 1

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,



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

Difficulty： Easy

Thinking 1

判断给定二叉树中是否存在从根节点到叶子节点沿路累加，总和为指定值的路径，用递归函数解决即可。递归基为某叶子节点的值等于sum；递归步骤为

return (hasPathSum(root -> left, sum - root -> val) || hasPathSum(root -> right, sum - root -> val))

递归过程从根节点开始，若该节点值不等于sum，则sum减去该节点值，再继续对左子树或右子树进行判断，一边成立即可。

Solution 1

class Solution {
public:
bool hasPathSum(TreeNode* root, int sum) {
if(!root) return false;
if(root -> val == sum && root -> left == NULL && root -> right == NULL) return true;
else return (hasPathSum(root -> left, sum - root -> val) || hasPathSum(root -> right, sum - root -> val));
}
};

Description 2

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,



return



Difficulty： Medium

Thinking 2

与上一题相同，也用递归的方法从根节点开始使sum沿路径递减，至最后一个叶子节点的值与sum相等则保存该路径。需要设置一个vector path存储路径，一个二维vector result保存所有路径。

将遍历过程遇到的节点存储到path中，若沿一条路径匹配成功，则将path存入result中，再从path尾部将节点pop，返回到上一个分叉节点；若匹配不成功，则直接将尾部节点pop，返回上一个分叉节点。

Solution 2

class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> result;
vector<int> path;
FindPath(root, sum, result, path);
return result;
}
private:
void FindPath(TreeNode* node, int sum, vector<vector<int>>& result, vector<int>& path){
if(!node) return;
path.push_back(node -> val);
if(!(node->right) && !(node -> left) && sum == node -> val){
result.push_back(path);
}
FindPath(node->left, sum - node -> val, result, path);
FindPath(node -> right, sum - node -> val, result, path);
path.pop_back();
}
};