leetcode 104. Maximum Depth of Binary Tree
- Maximum Depth of Binary Tree:题目链接
递归-DFS
AC:
1 2 3 4 5 6 7 8 9 10 11 12
| class Solution { public int maxDepth(TreeNode root) { if (root == null) { return 0; } int leftMaxDepth = maxDepth(root.left); int rightMaxDepth = maxDepth(root.right); return leftMaxDepth > rightMaxDepth ? leftMaxDepth+1 : rightMaxDepth + 1; } }
|
BFS
第一种写法:
在层次遍历的基础上,入队的时刻不止将当前结点入队,并且把当前结点的层数入队,用一个类(结构体)记录当前结点和其层数
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
| class Solution { class Node{ TreeNode node; int level; public Node(TreeNode node, int level){ this.node = node; this.level = level; } } public int maxDepth(TreeNode root) { if (root == null) { return 0; } int res = 0; Queue<Node> queue = new LinkedList<>(); queue.offer(new Node(root,1)); while (!queue.isEmpty()) { Node front = queue.poll(); if (front.level > res) { res = front.level; } if (front.node.left != null) { queue.offer(new Node(front.node.left, front.level + 1)); } if (front.node.right != null) { queue.offer(new Node(front.node.right, front.level + 1)); } } return res; } }
|
第二种写法:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
| class Solution { public int maxDepth(TreeNode root) { if (root == null) { return 0; } int res = 0; Queue<TreeNode> queue = new LinkedList<>(); queue.offer(root); while (!queue.isEmpty()) { res++; for (int size = queue.size(); size > 0; --size) { TreeNode front = queue.poll(); if (front.left != null) { queue.offer(front.left); } if (front.right != null) { queue.offer(front.right); } } } return res; } }
|