【刷题-LeetCode】199 Binary Tree Right Side View

1. Binary Tree Right Side View

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       <---

解1 层序遍历，每一层最后一个节点肯定是最右边的节点。在层序遍历时，用cur表示当前层的节点数，next表示下一层节点数，每次从队列里面出来一个，cur减1，当cur==0时，表明当前出队的节点已经是最右边的，然后令cur=next, next = 0；每进队一个节点，next增加1

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int>ans;
        if(root == NULL)return ans;

        queue<TreeNode*>q;
        q.push(root);
        int cur = 1, next = 0;
        while(!q.empty()){
            TreeNode* tmp = q.front();
            q.pop();
            cur--;
            if(tmp->left){
                q.push(tmp->left);
                next++;
            }
            if(tmp->right){
                q.push(tmp->right);
                next++;
            }
            if(cur == 0){
                ans.push_back(tmp->val);
                cur = next;
                next = 0;
            }
        }
        return ans;
    }
};

解2 dfs

class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        vector<int>ans;
        if(root == NULL)return ans;
        dfs(root, 0, ans);
        return ans;
    }
    void dfs(TreeNode* node, int level, vector<int>& ans){
        if(level == ans.size())ans.push_back(node->val);
        if(node->right)dfs(node->right, level+1, ans);
        if(node->left)dfs(node->left, level+1, ans);
    }
};