Question

剑指 Offer II 046. 二叉树的右侧视图

题目描述

给定一个二叉树的 根节点 root，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 1:

输入: [1,2,3,null,5,null,4]
输出: [1,3,4]

示例 2:

输入: [1,null,3]
输出: [1,3]

示例 3:

输入: []
输出: []

 

提示:

• 二叉树的节点个数的范围是 [0,100]
• -100 <= Node.val <= 100 

 

注意：本题与主站 199 题相同：https://leetcode.cn/problems/binary-tree-right-side-view/

解法

方法一

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def rightSideView(self, root: TreeNode) -> List[int]:
    ans = []
    if not root:
      return ans
    d = deque([root])
    while d:
      n = len(d)
      ans.append(d[0].val)
      for i in range(n):
        node = d.popleft()
        if node.right:
          d.append(node.right)
        if node.left:
          d.append(node.left)
    return ans

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  public List<Integer> rightSideView(TreeNode root) {
    List<Integer> ans = new ArrayList<>();
    if (root == null) {
      return ans;
    }
    Deque<TreeNode> q = new ArrayDeque<>();
    q.offer(root);
    while (!q.isEmpty()) {
      ans.add(q.peekFirst().val);
      for (int i = q.size(); i > 0; --i) {
        TreeNode node = q.poll();
        if (node.right != null) {
          q.offer(node.right);
        }
        if (node.left != null) {
          q.offer(node.left);
        }
      }
    }
    return ans;
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  vector<int> rightSideView(TreeNode* root) {
    vector<int> ans;
    if (!root) return ans;
    queue<TreeNode*> q;
    q.push(root);
    while (!q.empty()) {
      ans.push_back(q.front()->val);
      for (int i = q.size(); i > 0; --i) {
        auto node = q.front();
        q.pop();
        if (node->right) q.push(node->right);
        if (node->left) q.push(node->left);
      }
    }
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func rightSideView(root *TreeNode) []int {
  var ans []int
  if root == nil {
    return ans
  }
  q := []*TreeNode{root}
  for len(q) > 0 {
    ans = append(ans, q[0].Val)
    for i := len(q); i > 0; i-- {
      node := q[0]
      q = q[1:]
      if node.Right != nil {
        q = append(q, node.Right)
      }
      if node.Left != nil {
        q = append(q, node.Left)
      }
    }
  }
  return ans
}