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LC - E - 257. Binary Tree Paths (DFS)

Solutions in C++

Solution 1

This solution uses Depth-First Search (DFS) algorithm and recursion.

Be able to introduce and use new members to the Solution class (e.g., helper functions, variables, etc.)

Complexity Analysis:

• $T=O(n)$$T = O(n)$ where $n$$n$ is the number of nodes in the given binary tree

• $S=O(\log n)$$S = O(\log n)$ where $n$$n$ is the number of nodes in the given binary tree (Depth of recursion tree)

Solution:

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/**
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 * Definition for a binary tree node.
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 * struct TreeNode {
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 *     int val;
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 *     TreeNode *left;
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 *     TreeNode *right;
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 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
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 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
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 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
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 * };
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 */
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class Solution {
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public:
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    vector<string> binaryTreePaths(TreeNode* root) {
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        string s;   // will be used to build up each path
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        buildPath(root, s);
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        return paths;
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    }
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private: 
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    vector<string> paths;
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    // helper function
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    void buildPath(TreeNode *root, string s)
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    {
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        // if root does not exist (termination condition 1)
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        if (!root)
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            return;
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        // if root is a leaf (termination condition 2)
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        else if (root->left == root->right)  // (!root->left && !root->right)
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        {
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            s += to_string(root->val);
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            paths.push_back(s);
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        }
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        // if root is a not a leaf 
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        else
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        {
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            s = s + to_string(root->val) + "->";
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            // recursive step
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            buildPath(root->left, s);
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            buildPath(root->right, s);
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        }
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    }
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};