【leetcode】Path Sum I & II（middle）

最新推荐文章于 2025-08-04 23:59:01 发布

转载最新推荐文章于 2025-08-04 23:59:01 发布 · 67 阅读

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#数据结构与算法

本文探讨了二叉树中寻找特定和的根到叶路径的问题，包括判断是否存在这样的路径以及找出所有满足条件的路径。通过递归方法解决了两个核心问题，并提供了简洁的代码实现。

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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.

思路：树的题目，整体思路就是递归查找。三行解决。

bool hasPathSum(TreeNode *root, int sum) {
        if(root == NULL) return false;
        if(sum == root->val && (root->left == NULL) && (root->right == NULL)) return true; //和相等 且 是叶子结点

        return hasPathSum(root->left, sum - root->val) || hasPathSum(root->right, sum - root->val);
    }

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]
]

思路：找所有路径，也是用递归。发现满足的路径就压入。

vector<vector<int> > pathSum(TreeNode *root, int sum) {
        vector<vector<int> > ans;
        vector<int> tmpans;
        findSum(root, sum, tmpans, ans);
        return ans;

    }

    void findSum(TreeNode *root, int sum, vector<int> tmpans, vector<vector<int>> &ans)
    {
        if(root == NULL) return;
        if(sum == root->val && (root->left == NULL) && (root->right == NULL)) //满足条件 压入答案
        {
            tmpans.push_back(root->val);
            ans.push_back(tmpans);
        }
        else
        {
            tmpans.push_back(root->val);
        }
        findSum(root->left, sum - root->val, tmpans, ans);
        findSum(root->right, sum - root->val, tmpans, ans);
    }