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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)
        }
    }
    
    

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