题目：

二叉树的最小深度

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;
    }
}
View Code

 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;
View Code