[LeetCode October Challange] Day 22 - Minimum Depth of Binary Tree

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 2

Example 2:

Input: root = [2,null,3,null,4,null,5,null,6]
Output: 5

Constraints:

• The number of nodes in the tree is in the range [0, 10^5].
• -1000 <= Node.val <= 1000

---

Solution

recursive DFS

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if (!root) return 0;
        if (!root->left) return 1+minDepth(root->right);
        if (!root->right) return 1+minDepth(root->left);
        return 1+min(minDepth(root->left), minDepth(root->right));
    }
};

iterative DFS

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if (!root) return 0;
        stack<pair<TreeNode*, int>> s;
        int ans = INT_MAX;
        s.push({root, 1});
        while (!s.empty()) {
            pair<TreeNode*, int> node = s.top(); s.pop();
            if (!node.first->left && !node.first->right)
                ans = min(ans, node.second);
            if (node.first->left) s.push({node.first->left, node.second+1});
            if (node.first->right) s.push({node.first->right, node.second+1});
        }
        return ans;
    }
};

iterative BFS

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

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if (!root) return 0;
        queue<TreeNode*> q;
        int ans = 0;
        q.push(root);
        while (!q.empty()) {
            ++ans;
            int q_size = q.size();
            for (int i=0; i<q_size; ++i) {
                TreeNode* node = q.front();
                if (!node->left && !node->right) return ans;
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
                q.pop();
            }
        }
        return ans;
    }
};