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BinaryTreeRightSideView.java
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BinaryTreeRightSideView.java
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/*
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].
*/
import java.io.*;
import java.util.*;
public class BinaryTreeRightSideView {
public static class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) { val = x; }
}
public static List<Integer> rightSideView(TreeNode root) {
List<Integer> res = new ArrayList<Integer>();
if(root== null) return res;
Queue<TreeNode> q = new LinkedList<>();
q.offer(root);
while(!q.isEmpty()){
int s = q.size();
for(int i = 0; i< s; i++){
TreeNode a = q.poll();
if(a.left != null) q.offer(a.left);
if(a.right != null) q.offer(a.right);
if(i == s-1) res.add(a.val);
}
}
return res;
}
public static void main(String[] args) {
TreeNode root = new TreeNode(1);
TreeNode node2 = new TreeNode(2);
TreeNode node3 = new TreeNode(3);
TreeNode node4 = new TreeNode(4);
TreeNode node5 = new TreeNode(5);
TreeNode node6 = new TreeNode(6);
TreeNode node7 = new TreeNode(7);
TreeNode node8 = new TreeNode(8);
TreeNode node9 = new TreeNode(9);
TreeNode node10 = new TreeNode(10);
TreeNode node11 = new TreeNode(11);
root.left = node2;
root.right = node3;
node2.left = node4;
node2.right = node5;
node3.left = node6;
node3.right = node7;
node4.left = node8;
node4.right = node9;
node5.left = node10;
node5.right = node11;
node6.left = null;
node6.right = null;
node7.left = null;
node7.right = null;
node8.left = null;
node8.right = null;
node9.left = null;
node9.right = null;
node10.left = null;
node10.right = null;
node11.left = null;
node11.right = null;
List<Integer> res = rightSideView(root);
System.out.println("Right side view:");
for(int i = 0; i < res.size() - 1; i++){
System.out.print(res.get(i) + " -> ");
}
System.out.print(res.get(res.size()-1));
return;
}
}