Question

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。

# 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

/**
 * 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;
  }
}

/**
 * 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;
  }
};

/**
 * 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
}

/**
 * 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;
}

// 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 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。

# 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

/**
 * 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);
  }
}

/**
 * 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);
  }
};

/**
 * 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
}

/**
 * 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;
}

// 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;
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
  }
}