标签:link bfs art lis tco down assets bsp count
Given a n-ary 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.
For example, given a 3-ary
tree:
We should return its max depth, which is 3.
Note:
1000
.5000
.1 /* 2 // Definition for a Node. 3 class Node { 4 public int val; 5 public List<Node> children; 6 7 public Node() {} 8 9 public Node(int _val,List<Node> _children) { 10 val = _val; 11 children = _children; 12 } 13 }; 14 */ 15 class Solution { 16 public int maxDepth(Node root) { 17 if (root == null) { 18 return 0; 19 } 20 int max = 0; 21 for(Node child : root.children) { 22 max = Math.max(maxDepth(child), max); 23 } 24 return max + 1; 25 } 26 }
Solution 2. BFS(tracking the number of nodes that need to be visited at each level)
1 class Solution { 2 public int maxDepth(Node root) { 3 if(root == null) { 4 return 0; 5 } 6 int level = 0; 7 int currLevelCount = 1; 8 int nextLevelCount = 0; 9 Queue<Node> queue = new LinkedList<>(); 10 queue.add(root); 11 while(!queue.isEmpty()) { 12 Node currNode = queue.poll(); 13 currLevelCount--; 14 for(Node child : currNode.children) { 15 queue.add(child); 16 nextLevelCount++; 17 } 18 if(currLevelCount == 0) { 19 level++; 20 currLevelCount = nextLevelCount; 21 nextLevelCount = 0; 22 } 23 } 24 return level; 25 } 26 }
Follow up: Can you implement a BFS to a certain depth?
[LeetCode] 559. Maximum Depth of N-ary Tree
标签:link bfs art lis tco down assets bsp count
原文地址:https://www.cnblogs.com/lz87/p/9937842.html