本问题对应的 leetcode 原文链接：剑指 Offer 34. 二叉树中和为某一值的路径
问题描述
给你二叉树的根节点 root 和一个整数目标和 targetSum ，找出所有 从根节点到叶子节点 路径总和等于给定目标和的路径。
叶子节点 是指没有子节点的节点。
示例1 ：
输入：root = [5,4,8,11,null,13,4,7,2,null,null,5,1], targetSum = 22
输出：[[5,4,11,2],[5,8,4,5]]
示例2 ：
输入：root = [1,2,3], targetSum = 5
输出：[]
示例 3：
输入：root = [1,2], targetSum = 0
输出：[]
- 提示：
- 树中节点总数在范围
[0, 5000]内 -1000 <= Node.val <= 1000-1000 <= targetSum <= 1000
- 树中节点总数在范围
解题思路
视频讲解直达： 本题视频讲解
代码实现
class Solution {
List<List<Integer>> res;
List<Integer> tmp;
public List<List<Integer>> pathSum(TreeNode root, int target) {
res = new ArrayList<>();
tmp = new ArrayList<>();
dsf(root, target);
return res;
}
void dsf(TreeNode root, int target){
if(root == null){
return;
}
tmp.add(root.val);
target = target - root.val;
if(root.left == null && root.right == null && target == 0){
res.add(new ArrayList<>(tmp));
}
dsf(root.left, target);
dsf(root.right, target);
tmp.remove(tmp.size() - 1);
}
}
时间复杂度：O(n)
空间复杂度：O(n)
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution(object):
def pathSum(self, root, target):
"""
:type root: TreeNode
:type target: int
:rtype: List[List[int]]
"""
res = []
tmp = []
def dsf(root, target):
if not root:
return
tmp.append(root.val)
target -= root.val
if not root.left and not root.right and target == 0:
res.append(tmp[:])
dsf(root.left, target)
dsf(root.right, target)
tmp.pop()
dsf(root, target)
return res
C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int target) {
vector<vector<int>> res;
vector<int> tmp;
function<void(TreeNode*, int)> dfs = [&](TreeNode* root, int target) {
if (!root) {
return;
}
tmp.push_back(root->val);
target -= root->val;
if (!root->left && !root->right && target == 0) {
res.push_back(tmp);
}
dfs(root->left, target);
dfs(root->right, target);
tmp.pop_back();
};
dfs(root, target);
return res;
}
};
Go
/**
* Definition for a binary tree node.
* type TreeNode struct {
* Val int
* Left *TreeNode
* Right *TreeNode
* }
*/
func pathSum(root *TreeNode, target int) [][]int {
var res [][]int
var tmp []int
var dfs func(*TreeNode, int)
dfs = func(root *TreeNode, target int) {
if root == nil {
return
}
tmp = append(tmp, root.Val)
target -= root.Val
if root.Left == nil && root.Right == nil && target == 0 {
res = append(res, append([]int{}, tmp...))
}
dfs(root.Left, target)
dfs(root.Right, target)
tmp = tmp[:len(tmp)-1]
}
dfs(root, target)
return res
}
JS
/**
* @param {TreeNode} root
* @param {number} target
* @return {number[][]}
*/
var pathSum = function(root, target) {
var res = [];
var tmp = [];
dsf(root, target);
return res;
function dsf(root, target) {
if (root == null) {
return;
}
tmp.push(root.val);
target = target - root.val;
if (root.left == null && root.right == null && target == 0) {
res.push([...tmp]);
}
dsf(root.left, target);
dsf(root.right, target);
tmp.pop();
}
};
shuaidi