Question

199. 二叉树的右视图

English Version

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 

解法

方法一：BFS

使用 BFS 层序遍历二叉树，每层最后一个节点即为该层的右视图节点。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树节点个数。

# 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]:
    ans = []
    if root is None:
      return ans
    q = deque([root])
    while q:
      ans.append(q[-1].val)
      for _ in range(len(q)):
        node = q.popleft()
        if node.left:
          q.append(node.left)
        if node.right:
          q.append(node.right)
    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.peekLast().val);
      for (int n = q.size(); n > 0; --n) {
        TreeNode node = q.poll();
        if (node.left != null) {
          q.offer(node.left);
        }
        if (node.right != null) {
          q.offer(node.right);
        }
      }
    }
    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{{root}};
    while (!q.empty()) {
      ans.emplace_back(q.back()->val);
      for (int n = q.size(); n; --n) {
        TreeNode* node = q.front();
        q.pop();
        if (node->left) {
          q.push(node->left);
        }
        if (node->right) {
          q.push(node->right);
        }
      }
    }
    return ans;
  }
};

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

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function rightSideView(root: TreeNode | null): number[] {
  const ans = [];
  if (!root) {
    return ans;
  }
  const q = [root];
  while (q.length) {
    const n = q.length;
    ans.push(q[n - 1].val);
    for (let i = 0; i < n; ++i) {
      const { left, right } = q.shift();
      left && q.push(left);
      right && q.push(right);
    }
  }
  return ans;
}

// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//   TreeNode {
//     val,
//     left: None,
//     right: None
//   }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
use std::collections::VecDeque;
impl Solution {
  pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
    let mut res = vec![];
    if root.is_none() {
      return res;
    }
    let mut q = VecDeque::new();
    q.push_back(root);
    while !q.is_empty() {
      let n = q.len();
      res.push(q[n - 1].as_ref().unwrap().borrow().val);
      for _ in 0..n {
        if let Some(node) = q.pop_front().unwrap() {
          let mut node = node.borrow_mut();
          if node.left.is_some() {
            q.push_back(node.left.take());
          }
          if node.right.is_some() {
            q.push_back(node.right.take());
          }
        }
      }
    }
    res
  }
}

方法二：DFS

使用 DFS 深度优先遍历二叉树，每次先遍历右子树，再遍历左子树，这样每层第一个遍历到的节点即为该层的右视图节点。

时间复杂度 $O(n)$，空间复杂度 $O(n)$。其中 $n$ 为二叉树节点个数。

# 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]:
    def dfs(node, depth):
      if node is None:
        return
      if depth == len(ans):
        ans.append(node.val)
      dfs(node.right, depth + 1)
      dfs(node.left, depth + 1)

    ans = []
    dfs(root, 0)
    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 {
  private List<Integer> ans = new ArrayList<>();

  public List<Integer> rightSideView(TreeNode root) {
    dfs(root, 0);
    return ans;
  }

  private void dfs(TreeNode node, int depth) {
    if (node == null) {
      return;
    }
    if (depth == ans.size()) {
      ans.add(node.val);
    }
    dfs(node.right, depth + 1);
    dfs(node.left, depth + 1);
  }
}

/**
 * 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;
    function<void(TreeNode*, int)> dfs = [&](TreeNode* node, int depth) {
      if (!node) {
        return;
      }
      if (depth == ans.size()) {
        ans.emplace_back(node->val);
      }
      dfs(node->right, depth + 1);
      dfs(node->left, depth + 1);
    };
    dfs(root, 0);
    return ans;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func rightSideView(root *TreeNode) (ans []int) {
  var dfs func(*TreeNode, int)
  dfs = func(node *TreeNode, depth int) {
    if node == nil {
      return
    }
    if depth == len(ans) {
      ans = append(ans, node.Val)
    }
    dfs(node.Right, depth+1)
    dfs(node.Left, depth+1)
  }
  dfs(root, 0)
  return
}

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *   val: number
 *   left: TreeNode | null
 *   right: TreeNode | null
 *   constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 *   }
 * }
 */

function rightSideView(root: TreeNode | null): number[] {
  const ans = [];
  const dfs = (node: TreeNode | null, depth: number) => {
    if (!node) {
      return;
    }
    if (depth == ans.length) {
      ans.push(node.val);
    }
    dfs(node.right, depth + 1);
    dfs(node.left, depth + 1);
  };
  dfs(root, 0);
  return ans;
}