111. Minimum Depth of Binary Tree

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本文介绍两种求解二叉树最小深度的有效算法：一种是通过深度优先搜索(DFS)递归实现，另一种是利用层序遍历的方法。这两种算法都能有效找出从根节点到最近叶子节点的最短路径。

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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.

算法1：dfs递归的求解

class Solution {
public:
    int minDepth(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        if(root == NULL)return 0;
        int res = INT_MAX;
        dfs(root, 1, res);
        return res;
    }
    void dfs(TreeNode *root, int depth, int &res)
    {
        if(root->left == NULL && root->right == NULL && res > depth)
            {res = depth; return;}
        if(root->left)
            dfs(root->left, depth+1, res);
        if(root->right)
            dfs(root->right, depth+1, res);
    }
};

还有一种更直观的递归解法，分别求左右子树的最小深度，然后返回左右子树的最小深度中较小者+1

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

算法2：层序遍历二叉树，找到最先遍历到的叶子的层数就是树的最小高度

/**
 * Definition for binary tree
 * struct TreeNode {
 * int val;
 * TreeNode *left;
 * TreeNode *right;
 * TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode *root) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
         //层序遍历，碰到第一个叶子节点就停止,NULL作为每一层节点的分割标志
        if(root == NULL)return 0;
        int res = 0;
        queue<TreeNode*> Q;
        Q.push(root);
        Q.push(NULL);
        while(Q.empty() == false)
        {
            TreeNode *p = Q.front();
            Q.pop();
            if(p != NULL)
            {
                if(p->left)Q.push(p->left);
                if(p->right)Q.push(p->right);
                if(p->left == NULL && p->right == NULL)
                {
                    res++;
                    break;
                }
            }
            else 
            {
                res++;
                if(Q.empty() == false)Q.push(NULL);
            }
        }
        return res;
    }
};