[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.
求二叉树的最大深度问题用到深度优先搜索 Depth First Search，递归的完美应用，跟求二叉树的最小深度问题原理相同，参见代码如下：
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.right)));
}
}我们也可以使用层序遍历二叉树，然后计数总层数，即为二叉树的最大深度，注意 while 循环中的 for 循环的写法有个 trick，一定要将 q.size() 放在初始化里，而不能放在判断停止的条件中，因为q的大小是随时变化的，所以放停止条件中会出错，参见代码如下：
C++ 解法二：
class Solution {
public:
int maxDepth(TreeNode* root) {
if (!root) return 0;
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->left);
if (t->right) q.push(t->right);
}
}
return res;
}
};Java 解法二：
public class Solution {
public int maxDepth(TreeNode root) {
if (root == null) return 0;
int res = 0;
Queue<TreeNode> q = new LinkedList<>();
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
类似题目：
Balanced Binary Tree
Minimum Depth of Binary Tree
Maximum Depth of N-ary Tree
参考资料：
https://leetcode.com/problems/maximum-depth-of-binary-tree/
https://leetcode.com/problems/maximum-depth-of-binary-tree/discuss/34207/my-code-of-c-depth-first-search-and-breadth-first-search
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