199:Binary Tree Right Side View【树】【DFS】【BFS】
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题目链接:click~
/*题意:给出一个二叉树,假设你站在树的最右侧,求能看到的数*//** *思路:DFS遍历整棵树,先判断当前结点是否是最右侧的,接着遍历 * 右子树,再左子树。 */class Solution {public: void GetView(TreeNode *root, vector<int> &v, int level) {//level表示第几层 if(root == NULL) return; if(level == v.size()) //是最右侧的结点 v.push_back(root->val); GetView(root->right, v, level+1); GetView(root->left, v, level+1); } vector<int> rightSideView(TreeNode *root) { vector<int> v; if(root == NULL) return v; GetView(root, v, 0); return v; }};
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