111. Minimum Depth of Binary Tree

Description

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

Solutions

Solution 1: Recursion

The termination condition for recursion is when the current node is null, at which point return $0$. If one of the left or right subtrees of the current node is null, return the minimum depth of the non-null subtree plus $1$. If neither the left nor right subtree of the current node is null, return the smaller value of the minimum depths of the left and right subtrees plus $1$.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

Python3JavaC++GoTypeScriptRustJavaScriptC

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def minDepth(self, root: Optional[TreeNode]) -> int:
        if root is None:
            return 0
        if root.left is None:
            return 1 + self.minDepth(root.right)
        if root.right is None:
            return 1 + self.minDepth(root.left)
        return 1 + min(self.minDepth(root.left), self.minDepth(root.right))

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int minDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        if (root.left == null) {
            return 1 + minDepth(root.right);
        }
        if (root.right == null) {
            return 1 + minDepth(root.left);
        }
        return 1 + Math.min(minDepth(root.left), minDepth(root.right));
    }
}

/**
 * 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));
    }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func minDepth(root *TreeNode) int {
    if root == nil {
        return 0
    }
    if root.Left == nil {
        return 1 + minDepth(root.Right)
    }
    if root.Right == nil {
        return 1 + minDepth(root.Left)
    }
    return 1 + min(minDepth(root.Left), minDepth(root.Right))
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function minDepth(root: TreeNode | null): number {
    if (root == null) {
        return 0;
    }
    const { left, right } = root;
    if (left == null) {
        return 1 + minDepth(right);
    }
    if (right == null) {
        return 1 + minDepth(left);
    }
    return 1 + Math.min(minDepth(left), minDepth(right));
}