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.

Note: A leaf is a node with no children.

Example:

Given binary tree [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its minimum depth = 2.

algorithm thought

得到最浅的叶子节点的深度，直接用广度优先搜索即可，广度优先搜索有找的极值的性质

code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int minDepth(TreeNode* root) {
        if(root==NULL)
            return 0;
        queue<TreeNode*> qu;
        qu.push(root);
        TreeNode* pos=new TreeNode(-1);
        qu.push(pos);
        int res=1;
        while(!qu.empty()){
            TreeNode* fr=qu.front();
            qu.pop();
            if(fr==NULL)
                continue;
            if(fr==pos){
                qu.push(pos);
                res++;
                continue;
            }
            if(isleaf(fr))
                return res;
            qu.push(fr->left);
            qu.push(fr->right);
        }
        return res;
    }
    bool isleaf(TreeNode* root){
        return (root->right==NULL)&&(root->left==NULL);
    }
};

algorithm analysis

如果用深度优先搜索做这个题目，最后时间复杂度会是O(n)，因为会遍历所有节点，最后得到最浅的叶子节点。但是用广度优先搜索，只要找到第一个叶子节点就会返回，所以时间复杂度是降低很多的，不过应该也是O(n)