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); } }