题目：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.

DFS的题。

注意必须是root－leaf的长，即如果root只有右子树，那么最小深度就是右子树深度，不是0（左子树深度）。所以比较左右深度，只在子树不为0时才比较，否则就是INT最大VALUE。

第一次提交： min是global var，

class Solution {
    //必须是root－leaf的长
    int min;
    public int minDepth(TreeNode root) {
        if(root == null) return 0;
        return Math.min(Integer.MAX_VALUE, countDepth(root, 0));
    }
    
    private int countDepth(TreeNode root, int count){
        count++;
        if(root.left == null && root.right == null){
            return count;
        }
        int left = Integer.MAX_VALUE;
        int right = Integer.MAX_VALUE;
        if(root.left != null)  left = countDepth(root.left, count);
        if(root.right != null) right = countDepth(root.right, count);
        min = Math.min(left, right);
        return min;

    }
}

写完发现可以不需要int min。

class Solution {
    //必须是root－leaf的长
    public int minDepth(TreeNode root) {
        if(root == null) return 0;
        return Math.min(Integer.MAX_VALUE, countDepth(root, 0));
    }
    
    private int countDepth(TreeNode root, int count){
        count++;
        if(root.left == null && root.right == null){
            return count;
        }
        int left = Integer.MAX_VALUE;
        int right = Integer.MAX_VALUE;
        if(root.left != null)  left = countDepth(root.left, count);
        if(root.right != null) right = countDepth(root.right, count);
        return  Math.min(left, right);

    }
}