113 Path Sum II 路径总和 II
Description:
Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
Note:
A leaf is a node with no children.
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]
]
题目描述:
给定一个二叉树和一个目标和，找到所有从根节点到叶子节点路径总和等于给定目标和的路径。
说明:
叶子节点是指没有子节点的节点。
示例 :
给定如下二叉树，以及目标和 sum = 22，
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
返回:
[
[5,4,11,2],
[5,8,4,5]
]
思路:
回溯法
参考LeetCode #112 Path Sum 路径总和
时间复杂度O(n), 空间复杂度O(n)
代码:
C++:
/**
* 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>> pathSum(TreeNode* root, int sum)
{
vector<vector<int>> result;
vector<int> temp;
helper(result, root, sum, temp);
return result;
}
private:
void helper(vector<vector<int>> &result, TreeNode *root, int target, vector<int> &temp)
{
if (!root) return;
temp.push_back(root -> val);
if (root -> val == target and !root -> left and !root -> right) result.push_back(temp);
helper(result, root -> left, target - root -> val, temp);
helper(result, root -> right, target - root -> val, temp);
temp.pop_back();
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public List<List<Integer>> pathSum(TreeNode root, int sum) {
List<List<Integer>> result = new ArrayList<>();
List<Integer> temp = new ArrayList<>();
helper(result, root, sum, temp);
return result;
}
private void helper(List<List<Integer>> result, TreeNode root, int target, List<Integer> temp) {
if (root == null) return;
temp.add(root.val);
if (root.val == target && root.left == null && root.right == null) result.add(new ArrayList<>(temp));
helper(result, root.left, target - root.val, temp);
helper(result, root.right, target - root.val, temp);
temp.remove(temp.size() - 1);
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
def helper(root: TreeNode, target: int, path: List[int]) -> List[List[int]]:
if not root:
return []
path.append(root.val)
if not root.left and not root.right:
result = path[:]
path.pop()
if root.val == target:
return [result]
return []
target -= root.val
result = helper(root.left, target, path) + helper(root.right, target, path)
path.pop()
return result
return helper(root, sum, [])
