Algorithm Puzzles: Sum of Left Leaves

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Algorithm Puzzles everyday every week sometimes: Sum of Left Leaves

Puzzle

Puzzle from leetcode:

Given the root of a binary tree, return the sum of all left leaves.

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

Solution

It’s an easy puzzle can be resolved via dfs or bfs, here I use bfs:

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class Solution {
  public:
    int sumOfLeftLeaves(TreeNode* root) {
        int sum = 0;

        if (root == nullptr) {
            return sum;
        }

        std::queue<TreeNode*> q;
        std::queue<bool> isLeftNode;

        q.push(root);
        isLeftNode.emplace(false);

        while (!q.empty()) {
            auto cur = q.front();
            if (cur->left == nullptr && cur->right == nullptr &&
                isLeftNode.front()) {
                sum += cur->val;
            }
            q.pop();
            isLeftNode.pop();

            if (cur->left != nullptr) {
                q.push(cur->left);
                isLeftNode.emplace(true);
            }
            if (cur->right != nullptr) {
                q.push(cur->right);
                isLeftNode.emplace(false);
            }
        }

        return sum;
    }
};

TC should be O(n) and SC should be O(n) in worst case, where n is node number.