path-sum i & ii_小径真值验证-优快云博客

本文链接：https://blog.youkuaiyun.com/buki26/article/details/76080914

本文探讨了二叉树中寻找从根节点到叶子节点的路径，使得路径上所有节点值之和等于给定值的问题。介绍了两种情况：一种是判断是否存在这样的路径，另一种是找出所有满足条件的路径，并提供了相应的递归算法实现。

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i.
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree andsum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path5->4->11->2which sum is 22.

思路：递归。递归结束条件：左右子树都为空且这一次调用函数的sum 和当前root的val相等。
递归调用时，第二个参数用sum-root->val，这样后面直接判断给定的这个值和下一层节点是否相等就可以了。

代码：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    bool hasPathSum(TreeNode *root, int sum) {
        if(root==NULL)
            return false;
        if(root->left==NULL && root->right==NULL && sum-root->val==0)
            return true;
        return hasPathSum(root->left,sum-root->val) || hasPathSum(root->right,sum-root->val);
    }
};

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 andsum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]

思路同上，定义vector来存放答案

代码：

/**
 * 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) {
        vector<vector<int> > res;
        vector<int> temp;
        sumhelper(root,sum,temp,res);
        return res;
    }

    void sumhelper(TreeNode *root,int sum,vector<int> temp,vector<vector<int> >& res){
        if(root==NULL)
            return;
        temp.push_back(root->val);
        if(root->left==NULL && root->right==NULL && sum-root->val==NULL)
            res.push_back(temp);
        sumhelper(root->left,sum-root->val,temp,res);
        sumhelper(root->right,sum-root->val,temp,res);
    }
};