<LeetCode OJ> 199. Binary Tree Right Side View

Total Accepted: 40438 Total
Submissions: 117654 Difficulty: Medium

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.

For example:

Given the following binary tree,

   1            <---
 /   2     3         <---
 \       5     4       <---

You should return [1, 3, 4].

Credits:

Special thanks to @amrsaqr for adding this problem and creating all test cases.

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分析：

典型的广度优先遍历，每一层的最右边那个元素显然就是当层的右视可见节点！

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<int> rightSideView(TreeNode* root) {
        if(root==NULL)
            return result;
        queue<TreeNode *> que;
        que.push(root);
        //广度优先，总是压入每一层最右边的即可
        while(!que.empty())
        {
            int levelNum = que.size();//通过size来判断当层的结束
            for(int i=0; i<levelNum; i++)
            {
                if(que.front()->left != NULL) //先获取该节点下一层的左右子，再获取该节点的元素，因为一旦压入必弹出，所以先处理左右子
                    que.push(que.front()->left);
                if(que.front()->right != NULL)
                    que.push(que.front()->right);
                if(i== (levelNum-1))
                    result.push_back(que.front()->val);
                que.pop();
            }

        }
        return result;
    }
private:
     vector<int> result;
};

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原文地址：http://blog.csdn.net/ebowtang/article/details/51236061

原作者博客：http://blog.csdn.net/ebowtang

本博客LeetCode题解索引：http://blog.csdn.net/ebowtang/article/details/50668895