Java实现 LeetCode 199 二叉树的右视图

199. 二叉树的右视图

给定一棵二叉树,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。

示例:

输入: [1,2,3,null,5,null,4]
输出: [1, 3, 4]
解释:

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

PS:

1层序遍历

2递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution { 
  public List<Integer> rightSideView(TreeNode root) {
    List<Integer> res = new ArrayList<>();
        if(root == null) return res;
        Queue<TreeNode> queue = new LinkedList<>();
        queue.add(root);
        while(!queue.isEmpty()){
            int count = queue.size();
            while(count > 0){
                count--;
                TreeNode cur = queue.poll();
                if(count == 0){
                    //只有上一层的最后一个才能加入res
                    //如果右面有,就是右面
                    //右面没有,左面就是上一层的最后一个
                    res.add(cur.val);
                }
                //先加左面,先poll左面
                if(cur.left != null){
                    queue.add(cur.left);
                }
                if(cur.right != null){
                    queue.add(cur.right);
                }
            }
        }
        return res;
  }
 
}
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
   int[] max = {0};
  public List<Integer> rightSideView(TreeNode root) {
    List<Integer> res = new ArrayList<>();
    helper(res, root, 1);
    return res;
  }

  private void helper(List<Integer> res,TreeNode treeNode, int deep) {
    if (treeNode == null) {
      return;
    }
    if (deep > max[0]) {
      max[0] = deep;
      res.add(treeNode.val);
    }
    helper(res, treeNode.right, deep + 1);
    helper(res, treeNode.left, deep + 1);
  }
}
原文地址:https://www.cnblogs.com/a1439775520/p/13076065.html