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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：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：[]

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

我们可以使用广度优先搜索，定义一个队列 $\textit{q}$，将根节点放入队列中。每次从队列中取出当前层的所有节点，对于当前节点，我们先判断右子树是否存在，若存在则将右子树放入队列中；再判断左子树是否存在，若存在则将左子树放入队列中。这样每次取出队列中的第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。

Python3

# 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[0].val)
            for _ in range(len(q)):
                node = q.popleft()
                if node.right:
                    q.append(node.right)
                if node.left:
                    q.append(node.left)
        return ans

Java

/**
 * 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 k = q.size(); k > 0; --k) {
                TreeNode node = q.poll();
                if (node.right != null) {
                    q.offer(node.right);
                }
                if (node.left != null) {
                    q.offer(node.left);
                }
            }
        }
        return ans;
    }
}

C++

/**
 * 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.size()) {
            ans.push_back(q.front()->val);
            for (int k = q.size(); k; --k) {
                auto node = q.front();
                q.pop();
                if (node->right) {
                    q.push(node->right);
                }
                if (node->left) {
                    q.push(node->left);
                }
            }
        }
        return ans;
    }
};

Go

/**
 * 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[0].Val)
		for k := len(q); k > 0; k-- {
			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
}

TypeScript

/**
 * 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: number[] = [];
    if (!root) {
        return ans;
    }
    const q: TreeNode[] = [root];
    while (q.length > 0) {
        ans.push(q[0].val);
        const nq: TreeNode[] = [];
        for (const { left, right } of q) {
            if (right) {
                nq.push(right);
            }
            if (left) {
                nq.push(left);
            }
        }
        q.length = 0;
        q.push(...nq);
    }
    return ans;
}

Rust

// 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::cell::RefCell;
use std::collections::VecDeque;
use std::rc::Rc;
impl Solution {
    pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut ans = vec![];
        if root.is_none() {
            return ans;
        }
        let mut q = VecDeque::new();
        q.push_back(root);
        while !q.is_empty() {
            let k = q.len();
            ans.push(q[0].as_ref().unwrap().borrow().val);
            for _ in 0..k {
                if let Some(node) = q.pop_front().unwrap() {
                    let mut node = node.borrow_mut();
                    if node.right.is_some() {
                        q.push_back(node.right.take());
                    }
                    if node.left.is_some() {
                        q.push_back(node.left.take());
                    }
                }
            }
        }
        ans
    }
}

JavaScript

/**
 * 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) {
    const ans = [];
    if (!root) {
        return ans;
    }
    const q = [root];
    while (q.length > 0) {
        ans.push(q[0].val);
        const nq = [];
        for (const { left, right } of q) {
            if (right) {
                nq.push(right);
            }
            if (left) {
                nq.push(left);
            }
        }
        q.length = 0;
        q.push(...nq);
    }
    return ans;
};

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。

Python3

# 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(root: Optional[TreeNode], depth: int) -> None:
            if root is None:
                return
            if len(ans) == depth:
                ans.append(root.val)
            dfs(root.right, depth + 1)
            dfs(root.left, depth + 1)

        ans = []
        dfs(root, 0)
        return ans

Java

/**
 * 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 root, int depth) {
        if (root == null) {
            return;
        }
        if (ans.size() == depth) {
            ans.add(root.val);
        }
        dfs(root.right, depth + 1);
        dfs(root.left, depth + 1);
    }
}

C++

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

Go

/**
 * 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(root *TreeNode, depth int) {
		if root == nil {
			return
		}
		if len(ans) == depth {
			ans = append(ans, root.Val)
		}
		dfs(root.Right, depth+1)
		dfs(root.Left, depth+1)
	}
	dfs(root, 0)
	return
}

TypeScript

/**
 * 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 = (root: TreeNode | null, depth: number) => {
        if (!root) {
            return;
        }
        if (ans.length == depth) {
            ans.push(root.val);
        }
        dfs(root.right, depth + 1);
        dfs(root.left, depth + 1);
    };
    dfs(root, 0);
    return ans;
}

Rust

// 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::cell::RefCell;
use std::rc::Rc;
impl Solution {
    pub fn right_side_view(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut ans = Vec::new();
        fn dfs(node: Option<Rc<RefCell<TreeNode>>>, depth: usize, ans: &mut Vec<i32>) {
            if let Some(node_ref) = node {
                let node = node_ref.borrow();
                if ans.len() == depth {
                    ans.push(node.val);
                }
                dfs(node.right.clone(), depth + 1, ans);
                dfs(node.left.clone(), depth + 1, ans);
            }
        }
        dfs(root, 0, &mut ans);
        ans
    }
}

JavaScript

/**
 * 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) {
    const ans = [];
    const dfs = (root, depth) => {
        if (!root) {
            return;
        }
        if (ans.length == depth) {
            ans.push(root.val);
        }
        dfs(root.right, depth + 1);
        dfs(root.left, depth + 1);
    };
    dfs(root, 0);
    return ans;
};

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README.md

199. 二叉树的右视图

题目描述

解法

方法一：BFS

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

方法二：DFS

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

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199. 二叉树的右视图

题目描述

解法

方法一：BFS

Python3

Java

C++

Go

TypeScript

Rust

JavaScript

方法二：DFS

Python3

Java

C++

Go

TypeScript

Rust

JavaScript