Lintcode97-Maximum Depth of Binary Tree-Easy

97. Maximum Depth of Binary Tree

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.

Example

Example 1:

Input: tree = {}
Output: 0
Explanation: The height of empty tree is 0.

Example 2:

Input: tree = {1,2,3,#,#,4,5}
Output: 3	
Explanation: Like this:
   1
  /                 
 2   3                
    /                 
   4   5

注意：

Java中参数传递：（细读）

1> 基本数据类型作为参数传递时，基本类型作为参数传递时，是传递值的拷贝，无论你怎么改变这个拷贝，原值是不会改变的。

2> 对象作为参数传递时，是把对象在内存中的地址拷贝了一份传给了参数。

所以错误案例中，把基本数据类型int maxDepth作为参数，传给helper，只是拷贝了一份值给helper函数，所以无论helper里面maxDepth的值如何改变，return的值始终为1。

递归法代码（错误案例）：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: An integer
     */
    
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int maxDepth = 1;
        helper(root, 1, maxDepth);
        return maxDepth;
    }
    public void helper(TreeNode root, int curDepth, int maxDepth) {
        if (root == null) {
            return;
        }
        
        if (root.left == null && root.right == null) {
            if (curDepth > maxDepth) {
                maxDepth = curDepth;
            }
            return;
        }
        helper(root.left, curDepth + 1, maxDepth);
        helper(root.right, curDepth + 1, maxDepth);
    }
}

递归法代码（改正）：

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */

public class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: An integer
     */
    int maxDepth = 1;
    public int maxDepth(TreeNode root) {
        if (root == null) {
            return 0;
        }
        helper(root, 1);
        return maxDepth;
    }
    public void helper(TreeNode root, int curDepth) {
        if (root == null) {
            return;
        }
        
        if (root.left == null && root.right == null) {
            if (curDepth > maxDepth) {
                maxDepth = curDepth;
            }
            return;
        }
        helper(root.left, curDepth + 1);
        helper(root.right, curDepth + 1);
    }
}