[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。最大深度就是从根节点到叶节点的最长路径。
解法1: DFS
解法2: 层序遍历,总层数就是最大深度。
Java:DFS
public class Solution { public int maxDepth(TreeNode root) { return root == null ? 0 : (1 + Math.max(maxDepth(root.left), maxDepth(root.right))); } }
Java: Level Travesal
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; int n = q.size(); for (int i = 0; i < n; ++i) { TreeNode t = q.poll(); if (t.left != null) q.offer(t.left); if (t.right != null) q.offer(t.right); } } return res; } }
Python: DFS
class Solution: # @param root, a tree node # @return an integer def maxDepth(self, root): if root is None: return 0 else: return max(self.maxDepth(root.left), self.maxDepth(root.right)) + 1
C++: DFS
class Solution { public: int maxDepth(TreeNode* root) { if (!root) return 0; return 1 + max(maxDepth(root->left), maxDepth(root->right)); } };
C++:
class Solution { public: int maxDepth(TreeNode* root) { if (!root) return 0; int res = 0; queue<TreeNode*> q; q.push(root); while (!q.empty()) { ++res; int n = q.size(); for (int i = 0; i < n; ++i) { TreeNode *t = q.front(); q.pop(); if (t->left) q.push(t->left); if (t->right) q.push(t->right); } } return res; } };
类似题目:
[LeetCode] 366. Find Leaves of Binary Tree 找二叉树的叶节点
[LeetCode] 310. Minimum Height Trees 最小高度树
[LeetCode] 545. Boundary of Binary Tree 二叉树的边界
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