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 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.
思想： 先序遍历。若当前结点为空，返回 false； 不为空，则加上其值，若为叶子结点，则判断一次。
注意： 非路径和， 而是到叶子结点的路径和。例如： {1, 2} 1 返回: 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 and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
思路同上： 只是要记下路径。
/**
* 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;
}
};