[LeetCode#104, 111]Maximum Depth of Binary Tree, Minimum Depth of Binary Tree

The problem 1:

Given a binary tree, find its maximum depth.

The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.

My analysis:

The recursion solution for this problem is very elegant!
It include some skills we usually use in implenting recursive program for tree.
1. the base case for tree traversal is the leaf node's empty child(null), not the leaf node itself.

if (cur_root == null)
    return 0;

2. when we calculate the path from current node to the leaf node, we calculate it from bottom to surface. At each node, we plus one, and return it to previous level recursion.
3. We use Math.math() function to get the longest path for either sub-tree.

return Math.max(helper(cur_root.left), helper(cur_root.right)) + 1; 

My solution:

public class Solution {
    public int maxDepth(TreeNode root) {
        return helper(root);
    }
    
    private int helper(TreeNode cur_root) {
        
        if (cur_root == null)
            return 0;
        
        return Math.max(helper(cur_root.left), helper(cur_root.right)) + 1; 
    }
}

The problem2: 

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.

My analysis:

This problem is easy, but it differs a little with the previous problem: finding maximum path.
In this problem, we need to tackle with the "leaf node case" very carefully.
What's a leaf node?
A leaf node must has no left child or right child. We should mix this case with situation that a node has only one child. Some errors I have commited to solve the problem.
1. Directly return the minimum path among left-sub tree and right-sub tree. Apparently this is not right, what if a tree has only one left sub-tree with a length of 6?

return Math.min(left_min, right_min) + 1;

2. To fix the problem, I have tried to check if a node is a leaf node in each recursion. 

private int helper(TreeNode cur_root) {
    if (cur_root == null)
        return 0;
    if (cur_root.left == null && cur_root.right == null) 
        return 1;
        ...
    return Math.min(left_min, right_min) + 1;
}

But this still does not work, because we must keep the checking condition "if (cur_root == null)". (if a node has only one child, the empty child must be tackled). Then the solution would aslo suffer from the wrong judgement as previous one.

Fix:
How about check if the current node has one child or two, then tackle and return the value respectively.
1. check if the current node with only one child.
Note: iff current node is a leaf node, the following stataments can still tackle it.

if (left_min == 0)
    return right_min + 1;

if (right_min == 0)
    return left_min + 1;

2. if the current node with two children. 

return Math.min(left_min, right_min) + 1;

public class Solution {
    public int minDepth(TreeNode root) {
        
        if (root == null)
            return 0;
            
        return helper(root);
    }
    
    private int helper(TreeNode cur_root) {
        if (cur_root == null)
            return 0;
        
        int left_min = helper(cur_root.left);
        int right_min = helper(cur_root.right);
        
        if (left_min == 0)
            return right_min + 1;
        
        if (right_min == 0)
            return left_min + 1;
        
        return Math.min(left_min, right_min) + 1;
    }
}