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中等
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二叉树

515. 在每个树行中找最大值

English Version

题目描述

给定一棵二叉树的根节点 root ，请找出该二叉树中每一层的最大值。

 

示例1：

输入: root = [1,3,2,5,3,null,9]
输出: [1,3,9]

示例2：

输入: root = [1,2,3]
输出: [1,3]

 

提示：

  • 二叉树的节点个数的范围是 [0,104]
  • -231 <= Node.val <= 231 - 1

 

解法

方法一：BFS

BFS 找每一层最大的节点值。

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 largestValues(self, root: Optional[TreeNode]) -> List[int]:
        if root is None:
            return []
        q = deque([root])
        ans = []
        while q:
            t = -inf
            for _ in range(len(q)):
                node = q.popleft()
                t = max(t, node.val)
                if node.left:
                    q.append(node.left)
                if node.right:
                    q.append(node.right)
            ans.append(t)
        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> largestValues(TreeNode root) {
        List<Integer> ans = new ArrayList<>();
        if (root == null) {
            return ans;
        }
        Deque<TreeNode> q = new ArrayDeque<>();
        q.offer(root);
        while (!q.isEmpty()) {
            int t = q.peek().val;
            for (int i = q.size(); i > 0; --i) {
                TreeNode node = q.poll();
                t = Math.max(t, node.val);
                if (node.left != null) {
                    q.offer(node.left);
                }
                if (node.right != null) {
                    q.offer(node.right);
                }
            }
            ans.add(t);
        }
        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> largestValues(TreeNode* root) {
        if (!root) return {};
        queue<TreeNode*> q{{root}};
        vector<int> ans;
        while (!q.empty()) {
            int t = q.front()->val;
            for (int i = q.size(); i; --i) {
                TreeNode* node = q.front();
                t = max(t, node->val);
                q.pop();
                if (node->left) q.push(node->left);
                if (node->right) q.push(node->right);
            }
            ans.push_back(t);
        }
        return ans;
    }
};

Go

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

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 largestValues(root: TreeNode | null): number[] {
    const res: number[] = [];
    const queue: TreeNode[] = [];
    if (root) {
        queue.push(root);
    }
    while (queue.length) {
        const n = queue.length;
        let max = -Infinity;
        for (let i = 0; i < n; i++) {
            const { val, left, right } = queue.shift();
            max = Math.max(max, val);
            left && queue.push(left);
            right && queue.push(right);
        }
        res.push(max);
    }
    return res;
}

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 largest_values(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut res = Vec::new();
        let mut queue = VecDeque::new();
        if root.is_some() {
            queue.push_back(root.clone());
        }
        while !queue.is_empty() {
            let mut max = i32::MIN;
            for _ in 0..queue.len() {
                let node = queue.pop_front().unwrap();
                let node = node.as_ref().unwrap().borrow();
                max = max.max(node.val);
                if node.left.is_some() {
                    queue.push_back(node.left.clone());
                }
                if node.right.is_some() {
                    queue.push_back(node.right.clone());
                }
            }
            res.push(max);
        }
        res
    }
}

方法二：DFS

DFS 先序遍历，找每个深度最大的节点值。

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 largestValues(self, root: Optional[TreeNode]) -> List[int]:
        def dfs(root, curr):
            if root is None:
                return
            if curr == len(ans):
                ans.append(root.val)
            else:
                ans[curr] = max(ans[curr], root.val)
            dfs(root.left, curr + 1)
            dfs(root.right, curr + 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> largestValues(TreeNode root) {
        dfs(root, 0);
        return ans;
    }

    private void dfs(TreeNode root, int curr) {
        if (root == null) {
            return;
        }
        if (curr == ans.size()) {
            ans.add(root.val);
        } else {
            ans.set(curr, Math.max(ans.get(curr), root.val));
        }
        dfs(root.left, curr + 1);
        dfs(root.right, curr + 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> ans;

    vector<int> largestValues(TreeNode* root) {
        dfs(root, 0);
        return ans;
    }

    void dfs(TreeNode* root, int curr) {
        if (!root) return;
        if (curr == ans.size())
            ans.push_back(root->val);
        else
            ans[curr] = max(ans[curr], root->val);
        dfs(root->left, curr + 1);
        dfs(root->right, curr + 1);
    }
};

Go

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

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 largestValues(root: TreeNode | null): number[] {
    const res = [];
    const dfs = (root: TreeNode | null, depth: number) => {
        if (root == null) {
            return;
        }
        const { val, left, right } = root;
        if (res.length == depth) {
            res.push(val);
        } else {
            res[depth] = Math.max(res[depth], val);
        }
        dfs(left, depth + 1);
        dfs(right, depth + 1);
    };
    dfs(root, 0);
    return res;
}

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 {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, depth: usize, res: &mut Vec<i32>) {
        if root.is_none() {
            return;
        }
        let node = root.as_ref().unwrap().borrow();
        if res.len() == depth {
            res.push(node.val);
        } else {
            res[depth] = res[depth].max(node.val);
        }
        Self::dfs(&node.left, depth + 1, res);
        Self::dfs(&node.right, depth + 1, res);
    }

    pub fn largest_values(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
        let mut res = Vec::new();
        Self::dfs(&root, 0, &mut res);
        res
    }
}