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LC - E - 404. Sum of Left Leaves (DFS)

Solutions in C++

Solution 1

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

The strategy is to perform DFS and accumulate the values of the left leaves only.

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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    int sumOfLeftLeaves(TreeNode* root) {
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        int sum = 0;
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        helper(root, sum, 0);
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        return sum;
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    }
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private:
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    void helper(TreeNode *curr, int &sum, bool isLeft)
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    {
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        // terminating condition 1
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        if (curr == nullptr)
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            return;
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        // terminating condition 2
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        if (curr->left == curr->right)
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        {
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            sum += (curr->val * isLeft);
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            return;
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        }
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        // recursive step
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        helper(curr->left, sum, 1);
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        helper(curr->right, sum, 0);
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    }
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};

Solution 2

This solution does not use helper function or additional variable.

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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1
/**
2
 * Definition for a binary tree node.
3
 * struct TreeNode {
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 *     int val;
5
 *     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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    int sumOfLeftLeaves(TreeNode* root) {
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        if (!root)
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            return 0;
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        // if left node is a leaf
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        if (root->left && !root->left->left && !root->left->right)
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            return root->left->val + sumOfLeftLeaves(root->right);
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        // otherwise
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        return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
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    }
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};