113. 路径总和 II

原题

https://leetcode.cn/problems/path-sum-ii/description/

思路

深度优先搜索

复杂度

时间：O(n)
空间：O(n)

Python代码

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]:
        ans = []

        def dfs(root, path):
            if not root:
                return   
     
            path.append(root.val)
            if root.left is None and root.right is None and sum(path) == targetSum:
                ans.append(path[:])  
            dfs(root.left, path)
            dfs(root.right, path)
            path.pop()
        
        dfs(root, [])
        return ans 
        

Go代码

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func pathSum(root *TreeNode, targetSum int) [][]int {
	var ans [][]int
	var dfs func(*TreeNode, []int, int)
	dfs = func(root *TreeNode, path []int, curSum int) {
		if root == nil {
			return
		}
		path = append(path, root.Val)
		curSum += root.Val
		if root.Left == nil && root.Right == nil && curSum == targetSum {
			pathCopy := make([]int, len(path))
			copy(pathCopy, path)
			ans = append(ans, pathCopy)
		}
		dfs(root.Left, path, curSum)
		dfs(root.Right, path, curSum)
		path = path[:len(path)-1]
	}
	dfs(root, []int{}, 0)
	return ans
}