Leetcode之Path Sum II-优快云博客

本文探讨了一道经典的二叉树遍历问题，即如何找出所有从根节点到叶子节点且路径数值等于给定目标和的路径。通过深度优先搜索（DFS）策略，我们详细解析了递归算法的实现过程，包括如何处理叶节点判断、路径累加及回溯操作。

题目：

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]
]

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 {
private:
    vector<vector<int>> res;
public:
    void helper(TreeNode* root,int sum,vector<int>& v){
        if(!root)return;
        if(!root->left&&!root->right){
            if(sum!=root->val)return;
            v.push_back(root->val);
            res.push_back(v);
            v.pop_back();
        }
        int value=sum-root->val;
        v.push_back(root->val);
        if(root->left)helper(root->left,value,v);
        if(root->right)helper(root->right,value,v);
        v.pop_back();
    }
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<int> v;
        helper(root,sum,v);
        return res;
    }
};

想法：

还需要pop_back()，理解！