LeetCode—113.Path Sum II

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//\11134/\/\7251

Return:

[

  [5,4,11,2],

  [5,8,4,5]

]

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给定一棵二叉树，求问有没有一条从根节点到子节点额路径，各val值加起来等于sum。

与112题不同之处在于，要写明这条路径的所有节点的值。

采用回溯法，首先建立作为返回值的res数组，接着建立临时vector存储当前遍历路径的具体数值，若到达子节点，不符合条件，则清空临时vector；若符合条件，则将临时vector存入res数组中。

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/**

* 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> pathSum(TreeNode* root, int sum) {

        vector> res;

        vector temp;

        dfs(root, res, temp, sum);

        return res;

    }

    void dfs(TreeNode* root, vector>& res, vector& temp, int sum){

        if(!root) return;

        temp.push_back(root->val);

        int cur = sum - root->val;

        if(!root->left && !root->right && cur == 0){

            res.push_back(root->val);

            temp.pop_back();

            return;

        }

        if(!root->left && !root->right && cur != 0){

            temp.pop_back();

            return;

        }

        dfs(root->left, res, temp, sum-root->val);

        dfs(root->right, res, temp, sum-root->val);

        temp.pop_back();

        return;

    }

};