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.

Solution 2: BFS

Use a queue to implement breadth-first search, initially adding the root node to the queue.

Each time, take a node from the queue.

If this node is a leaf node, directly return the current depth.
If this node is not a leaf node, add all non-null child nodes of this node to the queue.

Continue to search the next layer of nodes until a leaf node is found.

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

/**
 * 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.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right

# dfs
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.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

from collections import deque

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

# bfs
class Solution:
    def minDepth(self, root: TreeNode) -> int:
        if not root:
            return 0
        
        queue = deque([(root, 1)])  # The queue holds tuples of (node, current_depth)
        
        while queue:
            current_node, depth = queue.popleft()
            
            # Check if the current node is a leaf node
            if not current_node.left and not current_node.right:
                return depth
            
            # Otherwise, add the children to the queue with incremented depth
            if current_node.left:
                queue.append((current_node.left, depth + 1))
            if current_node.right:
                queue.append((current_node.right, depth + 1))

# test case:
# Construct a binary tree: [3,9,20,None,None,15,7]
root = TreeNode(3)
root.left = TreeNode(9)
root.right = TreeNode(20)
root.right.left = TreeNode(15)
root.right.right = TreeNode(7)

sol = Solution()
print(sol.minDepth(root))  # Output: 2

/**
 * 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));
}

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number}
 */
var minDepth = function (root) {
    if (!root) {
        return 0;
    }
    if (!root.left) {
        return 1 + minDepth(root.right);
    }
    if (!root.right) {
        return 1 + minDepth(root.left);
    }
    return 1 + Math.min(minDepth(root.left), minDepth(root.right));
};

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>) -> i32 {
        if root.is_none() {
            return 0;
        }
        let node = root.as_ref().unwrap().borrow();
        if node.left.is_none() {
            return 1 + Self::dfs(&node.right);
        }
        if node.right.is_none() {
            return 1 + Self::dfs(&node.left);
        }
        1 + Self::dfs(&node.left).min(Self::dfs(&node.right))
    }

    pub fn min_depth(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
        Self::dfs(&root)
    }
}

111 - Minimum Depth of Binary Tree

111. Minimum Depth of Binary Tree

Description

Solutions

All Problems

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111. Minimum Depth of Binary Tree

Description

Solutions

All Problems

All Solutions