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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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