[112]. Path Sum,[113]. Path Sum II-优快云博客

本文链接：https://blog.youkuaiyun.com/u011934885/article/details/78588465

本文探讨了二叉树中寻找特定根到叶路径的问题，包括确定是否存在这样的路径及找出所有满足条件的路径。通过递归算法实现，介绍了两种方法，并提供了简洁高效的代码示例。

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[112]. Path Sum
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 and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \      \
        7    2      1

return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.

这题可以很简单的递归。深度优先搜索。

class Solution {
public:
    bool hasPathSum(TreeNode* root, int sum) {
        if (root == NULL) return false;
        if (root->left == NULL && root->right == NULL) return sum == root->val;
        if (root->left) {
            bool l = hasPathSum(root->left, sum - root->val);
            if (l) return true;
        } 
        if (root->right) {
            bool r = hasPathSum(root->right, sum - root->val);
            if (r) return true;
        }
        return false;
    }
};

[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.

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]
]

这题要把路径都保存下来。这也是树问题里很常见的一种。维持一个二维vector, 和一个一维vector，一维vector符合条件的时候添加进二维。至于一个一维如何在递归中生成多条路径，可以在传递的时候用值传递，也就是复制多处内存。或者就是每次添加完记得删除，如下代码所示。

其实有点像backtracking。

方法一：利用之前的path sum先判断有没有路径，有路径就添加。

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res;
        if (root == NULL) return res;
        vector<int> tempres;  
        helper2(root, sum, res, tempres);
        return res;
    }

    bool helper1(TreeNode* root, int sum) {
        if (root->left == NULL && root->right == NULL) {
            if (root->val == sum) {
                return true;
            }
        }
        if (root->left && helper1(root->left, sum - root->val)) return true;
        if (root->right && helper1(root->right, sum - root->val)) return true;
        return false;
    }

    void helper2(TreeNode* root, int sum, vector<vector<int>>& res, vector<int>& tempres) {
        if (helper1(root, sum)) {
            tempres.push_back(root->val);
            if (root->left == NULL && root->right == NULL) {
                res.push_back(tempres);
                tempres.pop_back();
                return;
            }
            if (root->left) {
                helper2(root->left, sum - root->val, res, tempres);
            }
            if (root->right) {
                helper2(root->right, sum - root->val, res, tempres);
            }
            tempres.pop_back();
        } 
    }

};

这里如果不想有pop_back(）的操作，可以直接在函数传递的时候传递vector tempres;

方法二：更加简洁的写法
先添加，再判断。

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> res;
        if (root == NULL) return res;
        vector<int> tempres;  
        helper(root, sum, res, tempres);
        return res;
    }

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

};