[Leetcode 113, Medium] Path sum II-优快云博客

本文链接：https://blog.youkuaiyun.com/u012212811/article/details/47269043

本文介绍了一种递归解决方案，用于在给定的二叉树中找到所有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的总和。包括了两种解决方案：递归和非递归方法。

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Problem:

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]
]

Analysis:

Solutions:

C++:

A recursive solution: (16 ms)

    void getPath(TreeNode *root, int sum, vector<vector<int> > &path, vector<int> &curPath)
    {
        if(root==NULL) return;
        
        if(root->left==NULL && root->right==NULL && root->val==sum){
            curPath.push_back(root->val);
            path.push_back(curPath);
            curPath.erase(curPath.end()-1, curPath.end());
        }else{
            curPath.push_back(root->val);
            
            if(root->left!=NULL)    getPath(root->left, sum-root->val, path, curPath);
            
            if(root->right!=NULL)   getPath(root->right, sum-root->val, path, curPath);
            
            curPath.erase(curPath.end()-1, curPath.end());
        }
    }
    
    vector<vector<int> > pathSum(TreeNode *root, int sum) {
        
        vector<vector<int> > path;
        if(root==NULL) return path;
        
        vector<int> curPath(1, root->val);
        if(root->left==NULL && root->right==NULL && root->val==sum) path.push_back(curPath);
        
        if(root->left!=NULL)    getPath(root->left, sum-root->val, path, curPath);
            
        if(root->right!=NULL)   getPath(root->right, sum-root->val, path, curPath);
        
        return path;
    }

A non-recursive solution: (16 ms)

    vector<vector<int>> pathSum(TreeNode* root, int sum) 
    {
        vector<vector<int>> results;
        if(root == NULL)  
            return results;

        stack<TreeNode *> node_stack;  
        TreeNode *p_cur = root;  
        TreeNode *visited = NULL;
        vector<int> result;
        while(p_cur || !node_stack.empty()) {  
            if(p_cur) {  
                sum -= p_cur->val;
                result.push_back(p_cur->val);
                if(p_cur->left == NULL && p_cur->right == NULL && sum == 0) {
                    results.push_back(result);
                }
                node_stack.push(p_cur);  
                p_cur = p_cur->left;  
            } else {  
                p_cur = node_stack.top();  
  
                if(p_cur->right == NULL || p_cur->right == visited) {  
                    sum += p_cur->val;  
                    result.erase(result.end() - 1);
                    node_stack.pop();  
                    visited = p_cur;  
                    p_cur = NULL;  
                } else  
                    p_cur = p_cur->right;  
            }  
        }
        
        return results;
    }