111. Minimum Depth of Binary Tree
https://leetcode.com/problems/minimum-depth-of-binary-tree/
Easy
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
Problem
Find the shallowest leaf node in the tree — what is its depth?
Approach
This problem is similar to 104 Maximum Depth of Binary Tree. This time, let’s use a different approach — solving it with an iterative (loop-based) method.
In each iteration, we check whether any node at the current level i has both Left and Right children as nil. If so, that’s the minimum depth — return it. If not, push the nodes of that level into a queue and check the next level.
- Declare a queue and push the root into it.
- Declare a depth variable to track the current depth.
- Start a while loop — continue as long as the queue length is not 0.
- Pop a node from the queue. Check if both left and right children are nil — if so, return the current depth. If not, push the left and right children into the queue and continue the loop from step 3.
Source Code
/**
* 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
}
depth := 0
queue := []*TreeNode{root}
for len(queue) > 0 {
depth++
length := len(queue)
for i := 0; i < length; i++ {
node := queue[0]
if node.Left == nil && node.Right == nil {
return depth
}
if node.Left != nil {
queue = append(queue, node.Left)
}
if node.Right != nil {
queue = append(queue, node.Right)
}
queue = queue[1:]
}
}
return depth
}
Closing
Do you have a better or simpler solution?
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