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.
Examples
Input: root = [3,9,20,null,null,15,7] Output: 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
Java Solution
Recursion
/**
* 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.right == null && root.left == null) return 1;
int minimumDepth = Integer.MAX_VALUE;
if(root.left != null) {
minimumDepth = Math.min(minDepth(root.left), minimumDepth);
}
if(root.right != null) {
minimumDepth = Math.min(minDepth(root.right), minimumDepth);
}
return minimumDepth + 1;
}
}
BFS Iteration
class Solution {
public int minDepth(TreeNode root) {
if(root == null) return 0;
Queue<TreeNode> queue = new ArrayDeque<>();
queue.offer(root);
int minDepth = 1;
while(!queue.isEmpty()) {
int size = queue.size();
for(int i = 0; i < size; i++) {
TreeNode current = queue.poll();
if(current != null && current.left == null && current.right == null) return minDepth;
if(current.left != null) queue.offer(current.left);
if(current.right != null) queue.offer(current.right);
}
minDepth += 1;
}
return -1;
}
}