# C++实现LeetCode(104.二叉树的最大深度)

[LeetCode] 104. Maximum Depth of Binary Tree 二叉树的最大深度 Given a binary tree, find its maximum depth.

## [LeetCode] 104. Maximum Depth of Binary Tree 二叉树的最大深度

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.

C++ 解法一：

```class Solution {
public:
int maxDepth(TreeNode* root) {
if (!root) return 0;
return 1 + max(maxDepth(root->left), maxDepth(root->right));
}
};```

java 解法一：

```public class Solution {
public int maxDepth(TreeNode root) {
return root == null ? 0 : (1 + Math.max(maxDepth(root.left), maxDepth(root.rightqKuyHqyntr)));
}
}```

C++ 解法二：

```class Solution {
public:
int maxDepth(TreeNode* root) {
if (!root) return 0http://www.cppcns.com;
int res = 0;
queue<TreeNode*> q{{root}};
while (!q.empty()) {
++res;
for (int i = q.size(); i > 0; --i) {
TreeNode *t = q.front(); q.pop();
if (t->left) q.push(t-&编程客栈gt;left);
if (t->right) q.push(t->right);
}
}
return res;
}
};```

Java 解法二：

```public www.cppcns.comclass Solution {
public int maxDepth(TreeNode root) {
if (http://www.cppcns.comroot == null) return 0;
int res = 0;
q.offer(root);
while (!q.isEmpty()) {
++res;
for (int i = q.size(); i > 0; --i) {
TreeNode t = q.poll();
if (t.left != null) q.offer(t.left);
if (t.right != null) q.offer(t.right);
}
}
return res;
}
}```

github 同步地址：

https://github.com/grandyang/leetcode/issues/104

Balanced Binary Tree

Minimum Depth of Binary Tree

Maximum Depth of N-ary Tree

https://leetcode.com/problems/maximum-depth-of-binary-tree/