[LeetCode] Path Sum II
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] ]
解题思路：
深度优先递归搜索。注意节点的值可能是负数。
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector> pathSum(TreeNode* root, int sum) {
vector> result;
vector resultItem;
helper(result, resultItem, root, sum);
return result;
}
void helper(vector>& result, vector resultItem, TreeNode* root, int leftSum){
if(root==NULL){
return;
}
if(root->left==NULL && root->right==NULL){ //root为叶子节点
if(leftSum==root->val){
resultItem.push_back(root->val);
result.push_back(resultItem);
}
}else{
resultItem.push_back(root->val);
leftSum = leftSum - root->val;
if(root->left!=NULL){
helper(result, resultItem, root->left, leftSum);
}
if(root->right!=NULL){
helper(result, resultItem, root->right, leftSum);
}
}
}
};
二次刷题（2015-08-04）
按引用传递速度更快
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector> pathSum(TreeNode* root, int sum) {
vector> result;
vector item;
helper(root, sum, item, result);
return result;
}
void helper(TreeNode* root, int left, vector& item, vector>& result){
if(root==NULL){
return;
}
item.push_back(root->val);
if(root->left==NULL && root->right==NULL && left == root->val){
result.push_back(item);
}
helper(root->left, left - root->val, item, result);
helper(root->right, left - root->val, item, result);
item.pop_back();
}
};
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