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.
For example:
Given the following binary tree,node
1 <--- / \ 2 3 <--- \ \ 5 4 <---
You should return [1, 3, 4].算法
给定一个二叉树,想象本身站在树的右边,返回从下到下你能看到的节点的值。ide
二叉树的层次遍历,每层按照从左向右的顺序依次访问节点,(每一层取最右边的结点)spa
树结点类.net
public class TreeNode { int val; TreeNode left; TreeNode right; TreeNode(int x) { val = x; } }
算法实现类code
public class Solution { public List<Integer> rightSideView(TreeNode root) { List<Integer> result = new LinkedList<>(); if (root != null) { Deque<TreeNode> deque = new LinkedList<>(); // 当前层的结点数 int current = 1; // 下一层的结点数 int next = 0; TreeNode node; deque.addLast(root); while (deque.size() > 0) { // 取第一个结点 node = deque.removeFirst(); current--; // 添加非空的左结点 if (node.left != null) { next++; deque.addLast(node.left); } // 添加非空的右结点 if (node.right != null) { next++; deque.addLast(node.right); } // 若是当前层已经处理完了 if (current == 0) { // 保存此层的最右一个结点值 result.add(node.val); // 设置下一层的元素个数 current = next; next = 0; } } } return result; } }