剑指offer——面试题39：二叉树的深度

力扣题目

Python

# 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 calculateDepth(self, root: Optional[TreeNode]) -> int:
        if not root:
            return 0
        return 1 + max(self.calculateDepth(root.left), self.calculateDepth(root.right))

Java

class Solution {
    
    public int calculateDepth(TreeNode root) {
        if (root == null) {
            return 0;
        } 

        return 1 + Math.max(calculateDepth(root.left), calculateDepth(root.right));
    }
}

20180906整理
##Solution1:
再本题中树的定义：若二叉树只有一个根节点，则此二叉树的深度为1.
迭代法，哈哈哈

/*
struct TreeNode {
	int val;
	struct TreeNode *left;
	struct TreeNode *right;
	TreeNode(int x) :
			val(x), left(NULL), right(NULL) {
	}
};*/
class Solution {
public:
    int TreeDepth(TreeNode* pRoot){
        if(pRoot == NULL)
            return 0;
        else {
            int res = GetTreeDepth(pRoot);
            return res;
        }
    }
    int GetTreeDepth(struct TreeNode *pRoot) {
        if(pRoot == NULL)
            return 0;
        return 1 + max(GetTreeDepth(pRoot->left), GetTreeDepth(pRoot->right));
    }
};