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发布于 2024-06-17 01:04:00 字数 10621 浏览 0 评论 0 收藏 0

513. 找树左下角的值

English Version

题目描述

给定一个二叉树的 根节点 root,请找出该二叉树的 最底层 最左边 节点的值。

假设二叉树中至少有一个节点。

 

示例 1:

输入: root = [2,1,3]
输出: 1

示例 2:

输入: [1,2,3,4,null,5,6,null,null,7]
输出: 7

 

提示:

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

解法

方法一:BFS

BFS 找最后一层第一个节点。

# 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 findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
    q = deque([root])
    ans = 0
    while q:
      ans = q[0].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 int findBottomLeftValue(TreeNode root) {
    Queue<TreeNode> q = new ArrayDeque<>();
    q.offer(root);
    int ans = 0;
    while (!q.isEmpty()) {
      ans = q.peek().val;
      for (int i = q.size(); i > 0; --i) {
        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:
  int findBottomLeftValue(TreeNode* root) {
    queue<TreeNode*> q{{root}};
    int ans = 0;
    while (!q.empty()) {
      ans = q.front()->val;
      for (int i = q.size(); i; --i) {
        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 findBottomLeftValue(root *TreeNode) int {
  q := []*TreeNode{root}
  ans := 0
  for len(q) > 0 {
    ans = q[0].Val
    for i := len(q); i > 0; i-- {
      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 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 findBottomLeftValue(root: TreeNode | null): number {
  let ans = 0;
  const q = [root];
  while (q.length) {
    ans = q[0].val;
    for (let i = q.length; i; --i) {
      const node = q.shift();
      if (node.left) {
        q.push(node.left);
      }
      if (node.right) {
        q.push(node.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 find_bottom_left_value(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
    let mut queue = VecDeque::new();
    queue.push_back(root);
    let mut res = 0;
    while !queue.is_empty() {
      res = queue.front().unwrap().as_ref().unwrap().borrow_mut().val;
      for _ in 0..queue.len() {
        let node = queue.pop_front().unwrap();
        let mut node = node.as_ref().unwrap().borrow_mut();
        if node.left.is_some() {
          queue.push_back(node.left.take());
        }
        if node.right.is_some() {
          queue.push_back(node.right.take());
        }
      }
    }
    res
  }
}

方法二:DFS

DFS 先序遍历,找深度最大的,且第一次被遍历到的节点。

# 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 findBottomLeftValue(self, root: Optional[TreeNode]) -> int:
    def dfs(root, curr):
      if root is None:
        return
      dfs(root.left, curr + 1)
      dfs(root.right, curr + 1)
      nonlocal ans, mx
      if mx < curr:
        mx = curr
        ans = root.val

    ans = mx = 0
    dfs(root, 1)
    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 int ans = 0;
  private int mx = 0;

  public int findBottomLeftValue(TreeNode root) {
    dfs(root, 1);
    return ans;
  }

  private void dfs(TreeNode root, int curr) {
    if (root == null) {
      return;
    }
    dfs(root.left, curr + 1);
    dfs(root.right, curr + 1);
    if (mx < curr) {
      mx = curr;
      ans = root.val;
    }
  }
}
/**
 * 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:
  int ans = 0;
  int mx = 0;
  int findBottomLeftValue(TreeNode* root) {
    dfs(root, 1);
    return ans;
  }

  void dfs(TreeNode* root, int curr) {
    if (!root) return;
    dfs(root->left, curr + 1);
    dfs(root->right, curr + 1);
    if (mx < curr) {
      mx = curr;
      ans = root->val;
    }
  }
};
/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func findBottomLeftValue(root *TreeNode) int {
  ans, mx := 0, 0
  var dfs func(*TreeNode, int)
  dfs = func(root *TreeNode, curr int) {
    if root == nil {
      return
    }
    dfs(root.Left, curr+1)
    dfs(root.Right, curr+1)
    if mx < curr {
      mx = curr
      ans = root.Val
    }
  }
  dfs(root, 1)
  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 findBottomLeftValue(root: TreeNode | null): number {
  let mx = 0;
  let ans = 0;

  function dfs(root, curr) {
    if (!root) {
      return;
    }
    dfs(root.left, curr + 1);
    dfs(root.right, curr + 1);
    if (mx < curr) {
      mx = curr;
      ans = root.val;
    }
  }
  dfs(root, 1);
  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 {
  fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, cur: i32, max: &mut i32, res: &mut i32) {
    if root.is_none() {
      return;
    }
    let root = root.as_ref().unwrap().borrow();
    Self::dfs(&root.left, cur + 1, max, res);
    Self::dfs(&root.right, cur + 1, max, res);
    if *max < cur {
      *max = cur;
      *res = root.val;
    }
  }

  pub fn find_bottom_left_value(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
    let mut max = 0;
    let mut res = 0;
    Self::dfs(&root, 1, &mut max, &mut res);
    res
  }
}

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