題目
輸入一棵二叉樹和一個整數,打印出二叉樹中節點值的和爲輸入整數的所有路徑。從樹的根節點開始往下一直到葉節點所經過的節點形成一條路徑。
示例:
給定如下二叉樹,以及目標和 sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
返回:
[
[5,4,11,2],
[5,8,4,5]
]
提示:
節點總數 <= 10000
代碼
Python
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
# 思路:
# 這是二次樹的搜索問題,採用回溯法:BFS+路徑記錄
# 複雜度:
# O(N)
class Solution:
def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]:
res, path = [], [] # 結果列表,路徑列表
def recur(root, sum):
if not root: return
path.append(root.val) # 路徑更新
sum -= root.val # 目標值更新
if sum == 0 and not root.left and not root.right:
# 注意上述三條件缺一不可
res.append(list(path)) # 正確路徑記錄,注意深拷貝
recur(root.left, sum) # 左子樹
recur(root.right, sum) # 右子樹
path.pop() # 路徑恢復,這是回溯法的精髓
recur(root, sum) # 根節點
return res
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>> res;
if (!root) return res;
vector<int> path;
recur(root, res, path, sum);
return res;
}
void recur(TreeNode* root, vector<vector<int>> &res, vector<int> path, int sum) {
if (!root) return;
sum -= root->val;
path.push_back(root->val);
if (sum == 0 && !root->left && !root->right) {
res.push_back(path);
return;
}
recur(root->left, res, path, sum);
recur(root->right, res, path, sum);
}
};