111. 二叉树的最小深度

# 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: TreeNode) -> int:
        if root is None:
            return 0
        if root.left is None and root.right is None:
            return 1
        l = self.minDepth(root.left)
        r = self.minDepth(root.right)
        # 如果左子树和右子树其中一个为空，那么需要返回比较大的那个子树的深度
        if root.left is None or root.right is None:
            return l + r + 1
        # 左右子树都不为空，返回最小深度+1即可
        return min(l, r) + 1

/**
 * 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 && root.right == null) return 1;
        int l = minDepth(root.left);
        int r = minDepth(root.right);
        if (root.left == null || root.right == null) return l + r + 1;
        return Math.min(l, r) + 1;
    }
}