标签：obj minimum test from mat The bfs += null
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.
Approach #1: C++. [recursive]
/**
* 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 == NULL) return 0;
return helper(root, 1);
}
private:
int helper(TreeNode* root, int depth) {
if (root->left != NULL && root->right != NULL)
return min(helper(root->left, depth+1), helper(root->right, depth+1));
else if (root->left != NULL)
return helper(root->left, depth+1);
else if (root->right != NULL)
return helper(root->right, depth+1);
return depth;
}
};
Approach #2: Java.
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public int minDepth(TreeNode root) {
if (root == null) return 0;
int left = minDepth(root.left);
int right = minDepth(root.right);
return (left == 0 || right == 0) ? left + right + 1 : Math.min(left, right) + 1;
}
}
Approach #3: Python. [BFS]
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
from Queue import *
class Solution(object):
def minDepth(self, root):
"""
:type root: TreeNode
:rtype: int
"""
if root == None:
return 0
q = Queue()
q.put(root)
ans = 0
while not q.empty():
ans += 1
k = q.qsize()
for i in range(k):
cur = q.get()
if cur.left:
q.put(cur.left)
if cur.right:
q.put(cur.right)
if cur.left == None and cur.right == None:
return ans
111. Minimum Depth of Binary Tree
标签：obj minimum test from mat The bfs += null
原文地址：https://www.cnblogs.com/ruruozhenhao/p/10092513.html
