leetcode 111 minimum depth of binary tree

problem 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.

problem analysis:

第一步，设计演算法：遍历一棵树的方法有两种，BFS and DFS，思考了下，BFS是一圈一圈的扩张出去，而这个problem是计算最小的深度，所以BFS可能更适合这个problem。

1.root is the layer 1

2.search out the node in layer 2 by layer1

3.check whether there is a leaf in layer 2: if true, return current layer; if false, continue iteration until finding a leaf.

第二步，设计data structure。在BFS中，优先考虑的data structure是queue，可是在c language中，并没有现成的queue container。那么how to solve this small problem。使用了两个数组，两个index，用来表示数组的长度，这样就实现了一个长度可变的数组。一个数组用来store父节点，另外一个数组用来store孩子节点。当一次iteration结束后，将孩子节点数组的内容移动到父节点数组，孩子节点数组清除为0，在code中，将孩子节点数组的index置为0表示数组当前值无效。

 /**
  * Definition for a binary tree node.
  * struct TreeNode {
  *     int val;
  *     struct TreeNode *left;
  *     struct TreeNode *right;
  * };
  */
 int minDepth(struct TreeNode* root) {
     //special case 1
     ;
     //special case 2
     ;

     int numOfParent, numOfChildren;
     ;
     ];
     ];

     //initialization
     parent[] = root;
     numOfParent = ;
     numOfChildren = ;

     //start iteration from root
     while(true)
     {
         //calculate children
         ;
         ; i<numOfParent; i++)
         {
             if(parent[i]->left) {children[counter]=parent[i]->left; counter++;}
             if(parent[i]->right) {children[counter]=parent[i]->right; counter++;}
         }
         //store the length of children in numOfChildren, children's level in layer
         numOfChildren = counter;
         layer++;
         //check whether there is a leaf in children
         ; k<numOfChildren; k++)
         {
             if( (children[k]->left==NULL) && (children[k]->right==NULL) ) return layer;
         }
         //preparation for next iteration
         numOfParent = numOfChildren;
         ; m<numOfChildren; m++)
         {
             parent[m] = children[m];
         }
         numOfChildren = ;

     }

 }