[leetcode] 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.

这道题和二叉树相关，找出二叉树中从根节点到叶节点的最短路径长度，题目难度为easy。

可以分别采用深度优先（DFS）和广度优先（BFS）两种方法进行计算。

所谓深度优先的方法即采用递归的方式从左子树和右子树中选出最短路径来确定最终的最短路径。具体代码：

class Solution {public:    int minDepth(TreeNode* root) {        if(!root) return 0;        else if(root->left && root->right) {            return min(minDepth(root->left), minDepth(root->right)) + 1;        }        else {            return minDepth(root->left) + minDepth(root->right) + 1;        }    }};

广度优先的方法需要按照层次序遍历二叉树，第一次遇到叶节点时最短路径也就确定了。由于是层次序遍历二叉树，这里需要用到队列。具体代码：

class Solution {public:    int minDepth(TreeNode* root) {        if(!root) return 0;                queue<TreeNode*> level;        int depth = 0;                level.push(root);        while(!level.empty()) {            depth++;            int sz = level.size();            for(int i=0; i<sz; i++) {                TreeNode* curNode = level.front();                if(!(curNode->left) && !(curNode->right)) return depth;                else {                    if(curNode->left) level.push(curNode->left);                    if(curNode->right) level.push(curNode->right);                }                level.pop();            }        }    }};