问题描述:给定一棵二叉树,想象自己站在它的右侧,按照从顶部到底部的顺序,返回从右侧所能看到的节点值。
思路:用DFS遍历二叉树,先访问右边的节点;访问过程中当当前深度大于访问过的最大深度时就更新最大深度,并将val压入res;
class Solution { public: Solution() { res = {}; max_floor = -1; } vector<int> res; int max_floor; void DFS(TreeNode* root, int now_floor) { if (root != NULL) { if (++now_floor > max_floor) { max_floor = now_floor; res.push_back(root->val); } DFS(root->right, now_floor); DFS(root->left, now_floor); } } vector<int> rightSideView(TreeNode* root) { DFS(root, -1); return res; } };
原文:https://www.cnblogs.com/airfy/p/12755363.html