LeetCode path Sum i ,ii递归和迭代解法_leetcode:path sum(i) 递归与非递归解法-优快云博客

本文链接：https://blog.youkuaiyun.com/taoyanqi8932/article/details/52006582

本文探讨了二叉树中寻找特定和的根到叶路径的问题。提供了两种算法：一种用于判断是否存在这样的路径，另一种则返回所有符合条件的路径集合。

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/********************************************************************************** 
* 
* 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.
************************************************************************/

// Date   : 2016.07.23
// Author : yqtao
// https://github.com/yqtaowhu
/**
 * 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 hasPathSum1(TreeNode *root, int sum) {
        if (root == NULL) return false;
        if (root->val == sum && root->left ==  NULL && root->right == NULL) return true;
        return hasPathSum(root->left, sum-root->val) || hasPathSum(root->right, sum-root->val);
    }

   bool  hasPathSum2(TreeNode *root, int sum){
    if(!n) return false;
    stack<TreeNode *> st;
    TreeNode *cur = root, *pre;
    while(cur || !st.empty()){
        while(cur){
            st.push(cur);
            sum -= cur->val;
            cur = cur->left;
        }
        cur = st.top();
        if(!cur->left && !cur->right && !sum) return true;
        if(cur->right && pre != cur->right) cur = cur->right;
        else{
            st.pop();
            sum += cur->val;
            pre = cur;
            cur = NULL;
        }
    }
    return false;
}  
};

********************************************************************************** 
* 
* 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]
* ]
* 
*               
**********************************************************************************/
class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int> > paths;
        vector<int> path;
        findPaths(root, sum, path, paths);
        return paths;  
    }
private:
    void findPaths(TreeNode* node, int sum, vector<int>& path, vector<vector<int> >& paths) {
        if (!node) return;
        path.push_back(node -> val);
        if (!(node -> left) && !(node -> right) && sum == node -> val)
            paths.push_back(path);
        findPaths(node -> left, sum - node -> val, path, paths);
        findPaths(node -> right, sum - node -> val, path, paths);
        path.pop_back();
    }
};