0199. 二叉树的右视图【中等】

• 1. 📝 题目描述
• 2. 🎯 s.1 - BFS

1. 📝 题目描述

• leetcode

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：root = [1,2,3,null,5,null,4]
输出：[1,3,4]

• 解释：
  • 

示例 2：

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

• 解释：
  • 

示例 3：

输入：root = [1,null,3]
输出：[1,3]

示例 4：

输入：root = []
输出：[]

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提示：

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

2. 🎯 s.1 - BFS

c

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int* rightSideView(struct TreeNode* root, int* returnSize) {
    *returnSize = 0;
    if (!root) return NULL;
    int* res = (int*)malloc(sizeof(int) * 10000);
    struct TreeNode** queue = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000);
    int front = 0, rear = 0;
    queue[rear++] = root;
    while (front < rear) {
        int size = rear - front;
        for (int i = 0; i < size; i++) {
            struct TreeNode* node = queue[front++];
            if (i == size - 1) res[(*returnSize)++] = node->val;
            if (node->left) queue[rear++] = node->left;
            if (node->right) queue[rear++] = node->right;
        }
    }
    free(queue);
    return res;
}

js

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {number[]}
 */
var rightSideView = function (root) {
  if (!root) return []
  const res = []
  const queue = [root]
  while (queue.length) {
    const size = queue.length
    for (let i = 0; i < size; i++) {
      const node = queue.shift()
      if (i === size - 1) res.push(node.val)
      if (node.left) queue.push(node.left)
      if (node.right) queue.push(node.right)
    }
  }
  return res
}

py

# 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: Optional[TreeNode]) -> List[int]:
        if not root:
            return []
        res = []
        queue = deque([root])
        while queue:
            size = len(queue)
            for i in range(size):
                node = queue.popleft()
                if i == size - 1:
                    res.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
        return res

• 时间复杂度：$O(n)$，其中 $n$ 是二叉树的节点数
• 空间复杂度：$O(n)$，队列最多存储一层的节点

算法思路：

• BFS 层序遍历，每层取最后一个节点的值（即该层最右侧的节点）