LeetCode算法题113：路径总和 II解析

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]
]

这个题需要用深度优先搜索，如果搜索到某个节点是叶子结点且之前的所有节点之和为sum，那就添加路径。

C++源代码：

/**
 * 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>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res;
        vector<int> out;
        DFS(root, out, res, sum);
        return res;
    }
    void DFS(TreeNode* root, vector<int>& out, vector<vector<int>>& res, int sum){
        if(!root) return;
        out.push_back(root->val);
        if(sum==root->val && !root->left && !root->right)
            res.push_back(out);
        DFS(root->left, out, res, sum-root->val);
        DFS(root->right, out, res, sum-root->val);
        out.pop_back();
    }
};

python3源代码：

# 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, sum):
        """
        :type root: TreeNode
        :type sum: int
        :rtype: List[List[int]]
        """
        l = []
        temp = []
        self.isPathSum(root, sum, l, temp)
        return l
    def isPathSum(self, root, sum, l, temp):
        if root==None:
            return
        temp.append(root.val)
        if root.left==None and root.right==None and sum==root.val:
            l.append(copy.deepcopy(temp))
        self.isPathSum(root.left, sum-root.val, l, temp)
        self.isPathSum(root.right, sum-root.val, l, temp)
        temp.pop()