Question

剑指 Offer II 045. 二叉树最底层最左边的值

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 

 

注意：本题与主站 513 题相同： https://leetcode.cn/problems/find-bottom-left-tree-value/

解法

方法一

# 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: TreeNode) -> int:
    q = deque([root])
    ans = -1
    while q:
      n = len(q)
      for i in range(n):
        node = q.popleft()
        if i == 0:
          ans = node.val
        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 = -1;
    while (!q.isEmpty()) {
      int n = q.size();
      for (int i = 0; i < n; i++) {
        TreeNode node = q.poll();
        if (i == 0) {
          ans = node.val;
        }
        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;
    q.push(root);
    int ans = -1;
    while (!q.empty()) {
      for (int i = 0, n = q.size(); i < n; ++i) {
        TreeNode* node = q.front();
        if (i == 0) ans = node->val;
        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 := -1
  for n := len(q); n > 0; n = len(q) {
    for i := 0; i < n; i++ {
      node := q[0]
      q = q[1:]
      if i == 0 {
        ans = node.Val
      }
      if node.Left != nil {
        q = append(q, node.Left)
      }
      if node.Right != nil {
        q = append(q, node.Right)
      }
    }
  }
  return ans
}