113. Path Sum II

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本文介绍了一种算法，用于在一个给定的二叉树中寻找所有从根节点到叶子节点的路径，其中路径上节点值之和等于指定的数值。通过递归方式实现，详细展示了如何保存和恢复路径信息。

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Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

For 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]
]

同理也是递归求解，只是要保存当前的结果，并且每次递归出来后要恢复递归前的结果，每当递归到叶子节点时就把当前结果保存下来。

/**
 * Definition for binary tree
 * 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) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<vector<int> > res;
        if(root == NULL)return res;
        vector<int> curres;
        curres.push_back(root->val);
        pathSumRecur(root, sum, curres, res);
        return res;
    }
    void pathSumRecur(TreeNode *root, int sum, vector<int> &curres, vector<vector<int> >&res)
    {
        if(root->left == NULL && root->right == NULL && root->val == sum)
        {
            res.push_back(curres);
            return;
        }
        if(root->left)
        {
            curres.push_back(root->left->val);
            pathSumRecur(root->left, sum - root->val, curres, res);
            curres.pop_back();
        }
        if(root->right)
        {
            curres.push_back(root->right->val);
            pathSumRecur(root->right, sum - root->val, curres, res);
            curres.pop_back();
        }
    }
};