标签：treenode col lin tree http 数字 root Plan binary
Given a binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can see ordered from top to bottom.
Example:
Input: [1,2,3,null,5,null,4] Output: [1, 3, 4] Explanation: 1 <--- / 2 3 <--- \ 5 4 <---
M1: BFS
level order traverse,，每次遍历完一层后，只把这一层最右的数字加入res中
time: O(n), space: O(N) -- N: most number of nodes in one level
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List<Integer> rightSideView(TreeNode root) { List<Integer> res = new ArrayList<>(); if(root == null) { return res; } Queue<TreeNode> q = new LinkedList<>(); q.offer(root); while(!q.isEmpty()) { List<Integer> tmp = new ArrayList<>(); int size = q.size(); for(int i = 0; i < size; i++) { TreeNode t = q.poll(); tmp.add(t.val); if(t.left != null) { q.offer(t.left); } if(t.right != null) { q.offer(t.right); } } res.add(tmp.get(tmp.size() - 1)); } return res; } }
M2: DFS
time: O(n), space: O(n)
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public List<Integer> rightSideView(TreeNode root) { List<Integer> res = new ArrayList<>(); if(root == null) { return res; } dfs(root, 0, res); return res; } public void dfs(TreeNode node, int curLevel, List<Integer> list) { if(node == null) { return; } if(curLevel == list.size()) { list.add(node.val); } dfs(node.right, curLevel + 1, list); dfs(node.left, curLevel + 1, list); } }
199. Binary Tree Right Side View - Medium
标签：treenode col lin tree http 数字 root Plan binary
原文地址：https://www.cnblogs.com/fatttcat/p/10201742.html
