LeetCode 104. Maximum Depth of Binary Tree--二叉樹高度--遞歸或迭代--C++,Python解法

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LeetCode題解專欄：LeetCode題解
我做的所有的LeetCode的題目都放在這個專欄裏，大部分題目Java和Python的解法都有。

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題目地址：Maximum Depth of Binary Tree - LeetCode

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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.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

3

/
9 20
/
15 7
return its depth = 3.

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這道題目意思很容易懂，就是求樹的高度，最簡單的做法是用遞歸。
Python解法如下：

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        if root==None:
            return 0
        return 1 +max(self.maxDepth(root.left),self.maxDepth(root.right))

遞歸的缺點是調用了很多次，使用迭代更快，使用迭代的話就要使用隊列了。

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def maxDepth(self, root: TreeNode) -> int:
        stack = []
        if root is not None:
            stack.append((1, root))
        
        depth = 0
        while stack != []:
            current_depth, root = stack.pop()
            if root is not None:
                depth = max(depth, current_depth)
                stack.append((current_depth + 1, root.left))
                stack.append((current_depth + 1, root.right))
        
        return depth

C++解法如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
  public:
  int maxDepth(TreeNode* root) {
        if(root == nullptr){
            return 0;
        }
        return max(maxDepth(root->left), maxDepth(root->right)) + 1;
  }
};