Question

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

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.

# 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));
}

// 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)
  }
}

/**
 * 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.
 * 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);
}

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.
# 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
    q = deque([root])
    ans = 0
    while 1:
      ans += 1
      for _ in range(len(q)):
        node = q.popleft()
        if node.left is None and node.right is None:
          return ans
        if node.left:
          q.append(node.left)
        if node.right:
          q.append(node.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;
    }
    Deque<TreeNode> q = new ArrayDeque<>();
    q.offer(root);
    int ans = 0;
    while (true) {
      ++ans;
      for (int n = q.size(); n > 0; n--) {
        TreeNode node = q.poll();
        if (node.left == null && node.right == null) {
          return ans;
        }
        if (node.left != null) {
          q.offer(node.left);
        }
        if (node.right != null) {
          q.offer(node.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;
    }
    queue<TreeNode*> q{{root}};
    int ans = 0;
    while (1) {
      ++ans;
      for (int n = q.size(); n; --n) {
        auto node = q.front();
        q.pop();
        if (!node->left && !node->right) {
          return ans;
        }
        if (node->left) {
          q.push(node->left);
        }
        if (node->right) {
          q.push(node->right);
        }
      }
    }
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func minDepth(root *TreeNode) (ans int) {
  if root == nil {
    return 0
  }
  q := []*TreeNode{root}
  for {
    ans++
    for n := len(q); n > 0; n-- {
      node := q[0]
      q = q[1:]
      if node.Left == nil && node.Right == nil {
        return
      }
      if node.Left != nil {
        q = append(q, node.Left)
      }
      if node.Right != nil {
        q = append(q, node.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) {
    return 0;
  }
  const q = [root];
  let ans = 0;
  while (1) {
    ++ans;
    for (let n = q.length; n; --n) {
      const node = q.shift();
      if (!node.left && !node.right) {
        return ans;
      }
      if (node.left) {
        q.push(node.left);
      }
      if (node.right) {
        q.push(node.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;
  }
  const q = [root];
  let ans = 0;
  while (1) {
    ++ans;
    for (let n = q.length; n; --n) {
      const node = q.shift();
      if (!node.left && !node.right) {
        return ans;
      }
      if (node.left) {
        q.push(node.left);
      }
      if (node.right) {
        q.push(node.right);
      }
    }
  }
};