LeetCode——Path Sum II

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本文介绍了一种算法，用于在给定的二叉树中找到所有从根节点到叶子节点的路径，使得这些路径上的节点值之和等于给定的目标和。通过递归遍历树的每个节点，算法能够高效地搜索满足条件的路径。

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]
]

» Solve this problem

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
private:
    void traversal(TreeNode *root, int sum, int cur, vector<vector<int> > &answer, vector<int> &temp) {
        temp.push_back(root->val);
        if (root->left == NULL && root->right == NULL) {
            if (root->val + cur == sum) {
                answer.push_back(temp);
            }
        }    
        else {
            if (root->left != NULL) {
                traversal(root->left, sum, root->val + cur, answer, temp);                
            }        
            if (root->right != NULL) {
                traversal(root->right, sum, root->val + cur, answer, temp);
            }
        }    
        temp.pop_back();
    }
    
public:
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int> > answer;
        vector<int> temp;
        if (root != NULL) {
            traversal(root, sum, 0, answer, temp);
        }
        return answer;
    }
};