Leetcode Binary Tree Right Side View

题目链接：https://leetcode.com/problems/binary-tree-right-side-view/

递归法解法：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;

public class Solution {
    public List<Integer> rightSideView(TreeNode root) {        
        Map<Integer,Integer> map = new HashMap<Integer,Integer>();
        breadthFirstSearch(root,map,0);
        
        List<Integer> list = new LinkedList<Integer>();
        
        for(Entry<Integer,Integer> entry:map.entrySet()){
            list.add(entry.getKey(),entry.getValue());
        }
        
        return list;
    }
    
    public void breadthFirstSearch(TreeNode root,Map<Integer,Integer> map,int depth){
        if(root == null){
            return;
        }
        
        if(!map.containsKey(depth)){
            map.put(depth,root.val);
        }
        
        breadthFirstSearch(root.right,map,depth+1);
        breadthFirstSearch(root.left,map,depth+1);
    }
}

 

 

 

遍历解法：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
import java.util.Map.Entry;

public class Solution {
    public List<Integer> rightSideView(TreeNode root) {

        List<Integer> list = new LinkedList<Integer>();
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        int count = 0;
        count++;
        if(root == null){
            return list;
        }
        queue.offer(root);
        
        while(true){
            int next = 0;
            while(count>0){
                TreeNode tmp = queue.poll();
                count--;
                if(count == 0){
                    list.add(tmp.val);
                }
                
                if(tmp.left != null){
                    queue.offer(tmp.left);
                    next++;
                }
                
                if(tmp.right != null){
                    queue.offer(tmp.right);
                    next++;
                }
            }
            
            if(next==0){
                break;
            }else{
                count = next;
            }
        }
        
        return list;
    }
}