康瑞部落

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.

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]
]

解题思路：

深度优先递归搜索。注意节点的值可能是负数。

/**
 * 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>> result;
        vector<int> resultItem;
        helper(result, resultItem, root, sum);
        return result;
    }
    
    void helper(vector<vector<int>>& result, vector<int> resultItem, TreeNode* root, int leftSum){
        if(root==NULL){
            return;
        }
        if(root->left==NULL && root->right==NULL){  //root为叶子节点
            if(leftSum==root->val){
                resultItem.push_back(root->val);
                result.push_back(resultItem);
            }
        }else{
            resultItem.push_back(root->val);
            leftSum = leftSum - root->val;
            if(root->left!=NULL){
                helper(result, resultItem, root->left, leftSum);
            }
            if(root->right!=NULL){
                helper(result, resultItem, root->right, leftSum);
            }
        } 
    }
};

二次刷题（2015-08-04）

按引用传递速度更快

/**
 * 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>> result;
        vector<int> item;
        helper(root, sum, item, result);
        return result;
    }
    
    void helper(TreeNode* root, int left, vector<int>& item, vector<vector<int>>& result){
        if(root==NULL){
            return;
        }
        item.push_back(root->val);
        if(root->left==NULL && root->right==NULL && left == root->val){
            result.push_back(item);
        }
        helper(root->left, left - root->val, item, result);
        helper(root->right, left - root->val, item, result);
        item.pop_back();
    }
};