Minimum Depth of Binary Tree LeetCode Solution

Problem – Minimum Depth of Binary Tree LeetCode Solution

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, 105].
  • -1000 <= Node.val <= 1000

Minimum Depth of Binary Tree LeetCode Solution in Java

public class Solution {
    public int minDepth(TreeNode root) {
        if(root == null) return 0;
        int left = minDepth(root.left);
        int right = minDepth(root.right);
        return (left == 0 || right == 0) ? left + right + 1: Math.min(left,right) + 1;
       
    }
}

Minimum Depth of Binary Tree LeetCode Solution in C++

int minDepth(TreeNode* root) {
    if (root == NULL) return 0;
    queue<TreeNode*> Q;
    Q.push(root);
    int i = 0;
    while (!Q.empty()) {
        i++;
        int k = Q.size();
        for (int j=0; j<k; j++) {
            TreeNode* rt = Q.front();
            if (rt->left) Q.push(rt->left);
            if (rt->right) Q.push(rt->right);
            Q.pop();
            if (rt->left==NULL && rt->right==NULL) return i;
        }
    }
    return -1; //For the compiler thing. The code never runs here.
}

Minimum Depth of Binary Tree LeetCode Solution in Python

def minDepth(self, root):
    if not root:
        return 0
    queue = collections.deque([(root, 1)])
    while queue:
        node, level = queue.popleft()
        if node:
            if not node.left and not node.right:
                return level
            else:
                queue.append((node.left, level+1))
                queue.append((node.right, level+1))
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