题目描述:
给定一个二叉树,找出其最小深度。
最小深度是从根节点到最近叶子节点的最短路径上的节点数量。
**说明:**叶子节点是指没有子节点的节点。
示例 1:
输入:root = [3,9,20,null,null,15,7]
输出:2
示例 2:
输入:root = [2,null,3,null,4,null,5,null,6]
输出:5
提示:
- 树中节点数的范围在
[0, 105]
内 -1000 <= Node.val <= 1000
题解:
/**
* 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 node) {
if (node == null) {
return 0;
}
if (node.left == null && node.right == null) {
return 1;
}
int minDepth = Integer.MAX_VALUE;
if (node.left != null) {
minDepth = Math.min(dfs(node.left), minDepth);
}
if (node.right != null) {
minDepth = Math.min(dfs(node.right), minDepth);
}
return minDepth + 1;
}
}