32. Path sum & Path sum II

Path sum

OJ: https://oj.leetcode.com/problems/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 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.

Idea: traverse in sequence. If the current node is empty, false is returned. If the current node is not empty, the value is added. If the node is a leaf node, it is judged once.

Note: The path is not the path and, but the path to the leaf node and. For example:{1, 2} 1 returns: false

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */bool judge(TreeNode *root, int curSum, int& sum) {    if(root == NULL) return false;    curSum += root->val;    if(!root->left && !root->right) return curSum == sum;    return judge(root->left, curSum, sum) || judge(root->right, curSum, sum);} class Solution {public:    bool hasPathSum(TreeNode *root, int sum) {        return judge(root, 0, sum);     }};

 

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 andsum = 22,

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

Return

[,], [,]
Same idea: Just write down the path.

/** * Definition for binary tree * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */void getPath(TreeNode *root, int curSum, int& sum, vector<vector<int> > &pathSet, vector<int> &path) {    if(root == NULL) return;    curSum += root->val;     path.push_back(root->val);    if(!root->left && !root->right && curSum == sum)          pathSet.push_back(path);    getPath(root->left, curSum, sum, pathSet, path);    getPath(root->right, curSum, sum, pathSet, path);    path.pop_back();}class Solution {public:    vector<vector<int> > pathSum(TreeNode *root, int sum) {        vector<vector<int> >pathSet;        vector<int> path;        getPath(root, 0, sum, pathSet, path);        return pathSet;    }};

 

32. Path sum & Path sum II