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[C++]LeetCode: 91 Path Sum II

2015-01-13

题目: Given a binary tree and a sum, find all root-to-leaf paths where each path & 039;s sum equals the given sum For example: Given the below binary tree and sum = 22,

题目:

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]
]

思路:构造一个辅助函数。辅助函数是一个深度搜索递归函数。递归终止条件,和Path Sum类似,当当前节点是叶子结点,并且当前节点值等于sum值,先把当前节点值push进tmp,再返回。如果当前节点非空,将当前结果压进tmp中,再接下来递归其左子树和右子树,寻找所有的可能组合。

Attention:

1. 注意只有存在左右孩子结点时,才递归调用辅助函数,搜索左右子树。

//需要判断是否有左右结点,没有就不递归调用了
            if(root->left) pathSum_helper(root->left, newSum, tmp, ret);
            if(root->right) pathSum_helper(root->right, newSum, tmp, ret);

2. 当判断当前节点符合迭代终止条件后,首先应先把当前节点值push进数组,再返回结果。(上一层调用中并没有push当前节点值)

/如果此时的节点就是叶子节点,并且符合sum条件,先push进tmp,再返回。
        if(!root->left && !root->right && root->val == sum)
        {
            tmp.push_back(root->val);
            ret.push_back(tmp);
            return;
        }
AC Code:

/**
 * 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 > pathSum(TreeNode *root, int sum) {
        vector > ret;
        vector tmp;
        if(root == NULL) return ret;
        pathSum_helper(root, sum, tmp, ret);
        return ret;
    }
    
private:
    void pathSum_helper(TreeNode* root, int sum, vector tmp, vector>& ret)
    {
        //如果此时的节点就是叶子节点,并且符合sum条件,先push进tmp,再返回。
        if(!root->left && !root->right && root->val == sum)
        {
            tmp.push_back(root->val);
            ret.push_back(tmp);
            return;
        }
       
        if(root != NULL)
        {
            tmp.push_back(root->val);
            int newSum = sum - root->val;
            //需要判断是否有左右结点,没有就不递归调用了
            if(root->left) pathSum_helper(root->left, newSum, tmp, ret);
            if(root->right) pathSum_helper(root->right, newSum, tmp, ret);
        }
        
        return;
    }
};



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