[leetcode] 111. Minimum Depth of Binary Tree

Description

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.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7
return its minimum depth = 2.

分析

题目的意思是：求二叉树的最小深度。

• 用了递归，取返回值中最小的深度+1，如果root的左右结点有一个为空，我们取左右结点的返回值最大值+1（可以带入上面的例子来思考）。
• 如果左右结点都非空，我们取左右结点的最小高度+1。
• -递归的代码很简洁。

C++实现

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if(!root){
            return 0;
        }
        if(!root->left||!root->right){
            return max(minDepth(root->left),minDepth(root->right))+1;
        }
        return min(minDepth(root->left),minDepth(root->right))+1;
    }
};

Python实现

下面是找规律的实现，如果找不到规律就完了。

# 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 minDepth(self, root: Optional[TreeNode]) -> int:
        if root is None:
            return 0
        if not root.left or not root.right:
            return max(self.minDepth(root.left),self.minDepth(root.right))+1
        return min(self.minDepth(root.left),self.minDepth(root.right))+1

如果找不到规律，那就中规中矩的分情况来写，对于每一个非叶子节点，我们只需要分别计算其左右子树的最小叶子节点深度。这样就将一个大问题转化为了小问题，可以递归地解决该问题。

# 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 minDepth(self, root: Optional[TreeNode]) -> int:
        if root is None:
            return 0
        if not root.left and not root.right:
            return 1
        min_depth = 2**20
        if root.left:
            min_depth = min(self.minDepth(root.left),min_depth)
        if root.right:
            min_depth = min(self.minDepth(root.right),min_depth)
        return min_depth+1

参考文献

[编程题]minimum-depth-of-binary-tree