113. Path Sum II

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最新推荐文章于 2025-09-06 17:22:57 发布

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文章标签： tree dfs leetcode

本文链接：https://blog.youkuaiyun.com/wilzxu/article/details/89222026

本文探讨了在二叉树中寻找所有从根节点到叶子节点路径的问题，这些路径的和等于给定值。通过深度优先搜索(DFS)算法，我们详细解释了如何有效地解决这个问题，并分析了其时间复杂度为O(n)，空间复杂度为O(h)。

113. Path Sum II

方法1：dfs
- Complexity
- 易错点：

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]
]

方法1：dfs

Complexity

Time complexity: O(n)
Space complexity: O(h)

易错点：

null只负责查空，leaf负责推入，这样做才能避免重复推和root为空以及即使sum=0也不算结果。

class Solution {
public:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        vector<vector<int>> result;
        vector<int> current;
        pathHelper(root, sum, result, current);
        return result;
    }
    
    void pathHelper(TreeNode* root, int sum, vector<vector<int>> & result, vector<int> & current){
        if (!root) return;

        current.push_back(root -> val);
        if (!root -> left && !root -> right && sum == root -> val){
            result.push_back(current); 
        }
        pathHelper(root -> left, sum - root ->val, result, current);
        pathHelper(root -> right, sum - root -> val, result, current);
        current.pop_back();
        
        return;
    }
};