1.问题描述
2.问题分析
本题采用回溯加先序遍历思想,思路大致是:
1.按照 “根、左、右” 的顺序,遍历树的所有节点
2.: 在先序遍历中,记录从根节点到当前节点的路径。当路径为 、
(1)根节点到叶节点形成的路径 且
(2)各节点值的和等于目标值 sum 时,将此路径加入结果列表。
3.代码实现
3.1C++代码
/**
* 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<vector<int>> VEC;
vector<int> vec;
vector<vector<int>> pathSum(TreeNode* root, int sum) {
recur(root,sum);
return VEC;
}
void recur(TreeNode *root,int target)
{
if(root==NULL)
return;
target-=root->val;
vec.push_back(root->val);
if(target==0&&root->left==NULL&&root->right==NULL)
{
VEC.push_back(vec);
}
recur(root->left,target);
recur(root->right,target);
vec.pop_back();
}
};
3.2Java代码
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
LinkedList<List<Integer>> res = new LinkedList<>();
LinkedList<Integer> path = new LinkedList<>();
public List<List<Integer>> pathSum(TreeNode root, int sum) {
recur(root, sum);
return res;
}
void recur(TreeNode root, int tar) {
if(root == null) return;
path.add(root.val);
tar -= root.val;
if(tar == 0 && root.left == null && root.right == null)
res.add(new LinkedList(path));
recur(root.left, tar);
recur(root.right, tar);
path.removeLast();
}
}