111. Minimum Depth of Binary Tree
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
DFS
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if(root==null){
return 0;
}
int left = minDepth(root.left);
int right = minDepth(root.right);
// 这一步很重要，注意只有一个分支的链表二叉树
if(left == 0 || right == 0){
return left+right+1;
}
return left<right?left+1:right+1;
}
}BFS
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if(root == null){
return 0;
}
Queue<TreeNode> queue = new LinkedList<>();
queue.offer(root);
int depth = 1;
while(!queue.isEmpty()){
int size = queue.size();
for(int i =0;i<size;i++){
TreeNode temp = queue.poll();
if(temp.left==null && temp.right==null){
return depth;
}
if(temp.left!=null){
queue.offer(temp.left);
}
if(temp.right!=null){
queue.offer(temp.right);
}
}
depth++;
}
return depth;
}
}