题目链接:二叉树的右视图

给定一个二叉树的 根节点 root,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。

方法一:层序遍历

/**
 * 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 == nullptr) return ans;
        queue<TreeNode*> q;
        q.push(root);
        int num = 0;
        while (q.size()) {
            int t = q.size();
            //int num;
            while (t--) {
                TreeNode* tmp = q.front();
                q.pop();
                num = tmp->val;
                if (tmp->left != nullptr) q.push(tmp->left);
                if (tmp->right != nullptr) q.push(tmp->right);
            }
            ans.push_back(num);
        }
        return ans;
    }
};

方法二:bfs

/**
 * 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) return ans;
        queue<TreeNode*> q;
        q.push(root);
        while (q.size()) {
            int size = q.size();
            ans.push_back(q.front()->val);
            for (int i = 0;i < size;i++) {
                TreeNode* node = q.front();
                q.pop();
                if (node->right) q.push(node->right);
                if (node->left) q.push(node->left);
            }
        }
        return ans;
    }
};

方法三:dfs

/**
 * 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:
    void dfs(TreeNode* root,int depth,vector<int>& ans) {
        if (root == nullptr) return ;
        if (depth == ans.size()) ans.push_back(root->val);
        dfs(root->right,depth+1,ans);
        dfs(root->left,depth+1,ans);
    }
    vector<int> rightSideView(TreeNode* root) {
        vector<int> ans;
        dfs(root,0,ans);
        return ans;
    }
};

标签: hot100, Medium, 二叉树, 层序遍历, dfs, bfs

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