LeetCode(111)Minimum Depth of Binary Tree

2024年1月18日 258次阅读

题目描述

Given a binary tree, find its minimum depth.

The minimum depth is the number of nodes along the shortest path from the root node down to the nearest leaf node.

Note: A leaf is a node with no children.

大意就是，计算二叉树的最小深度。

思路

递归思想。不同于求二叉树最大深度（也就是二叉树的深度定义），这里要考虑几种情况。

  • 左右子树都不为空
  • 左子树为空，右子树不为空
  • 左子树不为空，右子树为空

C++代码

// 90ms过大集合
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */
class Solution {
public:
     int minDepth(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(root!=NULL&&root->left!=NULL&&root->right!=NULL){
            int l=minDepth(root->left);
            int r=minDepth(root->right);
            return (l<r?l:r)+1;
        }else if(root!=NULL&&root->left==NULL){
            return 1+minDepth(root->right);
        }else if(root!=NULL&&root->right==NULL){
            return 1+minDepth(root->left);
        }else{
            return 0;
        }
    }
};

Python3代码

# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None

class Solution:
    def minDepth(self, root):
        """ :type root: TreeNode :rtype: int """
        if root!=None and root.left!=None and root.right!=None:
            ld = self.minDepth(root.left)
            rd = self.minDepth(root.right)
            return 1+(ld if ld < rd else rd)
        elif root!=None and root.left == None:
            return 1+self.minDepth(root.right)
        elif root!=None and root.right == None:
            return 1+self.minDepth(root.left)
        else:
            return 0

文章来源：
https://blog.csdn.net/feliciafay/article/details/18399489

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