LeetCode 199. 二叉树的右视图

 

思路#

方法：层序遍历#

保存层序遍历每一层的最后一个数字即可。

/**
 * 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> res;
        if(root == NULL)
            return res;

        //层序遍历
        queue<TreeNode*> q;
        q.push(root);
        TreeNode* t;
        while(!q.empty()) {
            int qsize = q.size();
            for(int i = 0; i < qsize; ++i) {
                t = q.front();
                //保存每一层的最后一个结点的值
                if(i == qsize-1) {
                    res.push_back(t->val);
                }
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
                q.pop();
            }
        }
        
        return res;
    }
};

复杂度分析：#

时间复杂度：O(n)

空间复杂度：O(n)