(待研究)111. Minimum Depth of Binary Tree

111. Minimum Depth of Binary Tree

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.

方法一、递归的做法

要求：此题正好和Maximum Depth of Binary Tree一题是相反的，即寻找二叉树的最小的深度值：从根节点到最近的叶子节点的距离。

结题思路：和找最大距离不同之处在于：找最小距离要注意(l<r)? l+1:r+1的区别应用，因为可能存在左右子树为空的情况，此时值就为0，但显然值是不为0的（只有当二叉树为空才为0），所以，在这里注意一下即可！

int minDepth(TreeNode *root) {          // Start typing your C/C++ solution below          // DO NOT write int main() function          if (root == NULL)              return 0;                        if (root->left == NULL && root->right == NULL)              return 1;                        int leftDepth = minDepth(root->left);          int rightDepth = minDepth(root->right);                    if (leftDepth == 0)              return rightDepth + 1;          else if (rightDepth == 0)              return leftDepth + 1;          else              return min(leftDepth, rightDepth) + 1;      } 

int minDepth(TreeNode* root) {        if(NULL == root)            return 0;        if(root->left == NULL)            return 1 + minDepth(root->right);        if(root->right == NULL)            return 1 + minDepth(root->left);        return 1 + min(minDepth(root->left),minDepth(root->right));    }

方法二、非递归的做法

level-order traversal and record current level depth, when meet a node which both child is null then return, no need to go farther

http://www.cnblogs.com/midhillzhou/p/6242613.html

或者见leetcode的top doscussion处