[LeetCode] (medium) 113. Path Sum II

最新推荐文章于 2021-06-11 03:09:39 发布

转载最新推荐文章于 2021-06-11 03:09:39 发布 · 171 阅读

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本文详细解析了LeetCode上路径总和II问题的三种解决方案：递归、迭代及后序遍历方法。通过实例讲解如何在二叉树中寻找所有从根节点到叶子节点的路径，使路径上的节点值之和等于指定值。

https://leetcode.com/problems/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.

Note: A leaf is a node with no children.

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]
]

递归版：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<vector<int>> result;
    
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        if(root == NULL) return result;
        vector<int> inter_num;
        dfs(root, inter_num, sum);
        return result;
    }
    
    void dfs(TreeNode *cur_node, vector<int> &inter_num, int cnt){
        inter_num.push_back(cur_node->val);
        cnt -= cur_node->val;
        if(cur_node->left == NULL && cur_node->right == NULL){
            if(cnt == 0) result.push_back(inter_num);
        }else{
            if(cur_node->left != NULL) dfs(cur_node->left, inter_num, cnt);
            if(cur_node->right != NULL) dfs(cur_node->right, inter_num, cnt);
        }
        inter_num.pop_back();
    }
    
    
};

迭代版：

/**
 * Definition for a binary tree node.
 * 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<TreeNode*> stack;
        vector<vector<int>> ans;
        vector<int> path;
        TreeNode *pre=NULL,*cur=root;
        int check=0;
        while(cur!=NULL || stack.size()!=0){
            while(cur!=NULL){
                stack.push_back(cur);
                path.push_back(cur->val);
                check+=cur->val;
                cur=cur->left;
            }
            cur=stack.back();
            if(cur->left==NULL && cur->right==NULL && check==sum){
                ans.push_back(path);
                
            }
            if(cur->right!=NULL && cur->right!=pre){
                cur=cur->right;
                continue;
            }
            
            pre=cur;
            stack.pop_back();
            path.pop_back();
            check-=cur->val;
            cur=NULL;
            
        }
        return ans;
    }
};

后序遍历版：

/**
 * Definition for a binary tree node.
 * 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>> result;
        vector<int> inter_nums;
        int inter_cnt = 0;
        stack<TreeNode*> S;
        
        if(root == NULL) return result;
        
        S.push(root);
        S.push(root);
        TreeNode *cur = root;
        
        while(!S.empty()){
            cur = S.top();  S.pop();
            if(!S.empty() && cur == S.top()){
                // cout << "F: " << cur->val << endl;
                inter_nums.push_back(cur->val);
                inter_cnt += cur->val;
                if(cur->left == NULL && cur->right == NULL){
                    if(inter_cnt == sum) result.push_back(inter_nums);
                }else{
                    if(cur->right != NULL){
                        S.push(cur->right);
                        S.push(cur->right);
                    }
                    if(cur->left != NULL){
                        S.push(cur->left);
                        S.push(cur->left);
                    }
                    
                }
            }else{
                // cout << "S: " << cur->val << endl;
                inter_cnt -= cur->val;
                inter_nums.pop_back();
            }
        }
        
        return result;
        
    }
};