题目：
给定一个二叉树，找出其最小深度。
二叉树的最小深度为根节点到最近叶子节点的距离。
样例
给出一棵如下的二叉树:
1
/ \
2 3
/ \
4 5
这个二叉树的最小深度为 2
解题：
递归求解
Java程序：
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: An integer.
*/
public int minDepth(TreeNode root) {
// write your code here
int res = 0; return depth(res,root);
}
public int depth(int res,TreeNode root){
if(root==null)
return 0;
if(root.left==null && root.right==null)
return res+1;
if(root.left!=null && root.right==null)
return res=depth(res,root.left)+1;
if(root.left==null && root.right!=null)
return res=depth(res,root.right)+1;
int res1 = depth(res,root.left)+1;
int res2 = depth(res,root.right)+1;
res = res1>res2?res2:res1;
return res;
}
}
总耗时: 2455 ms
Python程序：
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
"""
class Solution:
"""
@param root: The root of binary tree.
@return: An integer
"""
def minDepth(self, root):
# write your code here
if root==None:
return 0
if root.left==None and root.right!=None:
return self.minDepth(root.right) + 1
if root.left!=None and root.right==None:
return self.minDepth(root.left) + 1
return min(self.minDepth(root.right),self.minDepth(root.left)) + 1;