Leetcode:binary Tree Right Side View

Binary Tree Right Side View


Total Accepted: 44458 Total Submissions: 125991 Difficulty: Medium

Given A binary tree, imagine yourself standing on the right side of it, return the values of the nodes you can SE E 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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Hide TagsTree Depth-first Search Breadth-first SearchHide Similar Problems(M) populating Next right pointers in each Node

Ideas:

Hierarchical traversal.

C + + code:

/** * 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) {    vector<int> ret;    if (!root) return ret;    Queue<treenode*> que;    Que.push (root);    while (!que.empty ()) {    int size = Que.size ();    for (int i = 0; i < size; i++) {    TreeNode *tmp = Que.front ();    Que.pop ();    if (i = = 0) ret.push_back (tmp->val);    if (tmp->right) Que.push (tmp->right);    if (tmp->left) Que.push (tmp->left);    }    }    return ret;    }};

Java code:

/** * Definition for a binary tree node.  * public class TreeNode {*     int val, *     TreeNode left, *     TreeNode right; *     TreeNode (int x) {val = x;} *} */public class Solution {public    list<integer> Rightsideview (TreeNode root) {                list<integer> result = new arraylist<integer> ();        if (root = null) return result;        queue<treenode> queue = new linkedlist<treenode> ();        Queue.offer (root);        while (!queue.isempty ()) {            int size = Queue.size ();            for (int i=0;i<size;i++) {                TreeNode tmp = Queue.poll ();                if (i==0) Result.add (tmp.val);                if (tmp.right! = null) Queue.offer (tmp.right);                if (tmp.left! = null) Queue.offer (tmp.left);}        }        return result;}    }

Leetcode:binary Tree right Side View