LeetCode 104 Maximum Depth of Binary Tree
難度:Easynode
題目描述:
找到一顆二叉樹的最深深度。
Given a binary tree, find its maximum depth.code
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.遞歸
Note: A leaf is a node with no children.ip
Example:leetcode
Given binary tree [3,9,20,null,null,15,7]
,get
3 / \ 9 20 / \ 15 7
return its depth = 3.it
解題思路:
Solution 1:
解二叉樹的題目通常若是難度爲easy,則要求iterative和recursive都會寫,二叉樹的最深深度是左子樹和右子樹的Max深度,根據這一特性,咱們自底向上,時間複雜度O(n), Space O(1), 但由於是遞歸,因此會佔用stack,recursive的寫法以下:io
public int maxDepth(TreeNode root) { if (root == null) { return 0; } //get the left and right depth int leftDep = maxDepth(root.left); int rightDep = maxDepth(root.right); return 1 + Math.max(leftDep, rightDep); }
解題思路:
Solution 2:
使用BFS方法iterative使用一個Queue,每往下一層都加1,最後拿到最深深度。
時間:O(n), 空間 O(n)List
public int maxDepth(TreeNode root) { //iterative if (root == null) return 0; Queue<TreeNode> q = new LinkedList<>(); q.offer(root); int dep = 0; while(!q.isEmpty()) { dep++; int size = q.size(); for(int i = 0; i < size; i++) { TreeNode now = q.poll(); if (now.left != null) q.offer(now.left); if (now.right != null) q.offer(now.right); } } return dep; }