README_EN.md

111. Minimum Depth of Binary Tree

中文文档

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

Python3

# 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: TreeNode) -> int:
        def dfs(root):
            if root is None:
                return 0
            if root.left is None:
                return 1 + dfs(root.right)
            if root.right is None:
                return 1 + dfs(root.left)
            return 1 + min(dfs(root.left), dfs(root.right))

        return dfs(root)

Java

/**
 * 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) {
        return dfs(root);
    }

    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        if (root.left == null) {
            return 1 + dfs(root.right);
        }
        if (root.right == null) {
            return 1 + dfs(root.left);
        }
        return 1 + Math.min(dfs(root.left), dfs(root.right));
    }
}

C++

/**
 * 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) {
        return dfs(root);
    }

    int dfs(TreeNode* root) {
        if (!root) return 0;
        if (!root->left) return 1 + dfs(root->right);
        if (!root->right) return 1 + dfs(root->left);
        return 1 + min(dfs(root->left), dfs(root->right));
    }
};

Go

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

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

JavaScript

/**
 * 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) {
    function dfs(root) {
        if (!root) return 0;
        if (!root.left) return 1 + dfs(root.right);
        if (!root.right) return 1 + dfs(root.left);
        return 1 + Math.min(dfs(root.left), dfs(root.right));
    }
    return dfs(root);
};

TypeScript

/**
 * 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));
}

Rust

// 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)
    }
}

C

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

#define min(a,b) (((a) < (b)) ? (a) : (b))

int minDepth(struct TreeNode *root) {
    if (!root) {
        return 0;
    }
    if (!root->left) {
        return 1 + minDepth(root->right);
    }
    if (!root->right) {
        return 1 + minDepth(root->left);
    }
    int left = minDepth(root->left);
    int right = minDepth(root->right);
    return 1 + min(left, right);
}

...