letecode [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.
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
/ \
9 20
/ \
15 7
return its minimum depth = 2.
题目大意:
给定二叉树,计算它从根节点到某个叶节点的最小路径(即节点数)。
理 解 :
自己的思想是:从根节点开始,往下访问子节点,找到第一个叶节点,作为当前最小深度。后面遍历其他子树的时候,大于最小深度的子树就剪枝,小于就更新最小深度。实现的时候有点很奇怪的问题,暂时没调试出来。
看了别人的思想。这个真的很棒。用队列辅助层次遍历,当访问的节点的左右子树均为空时,则当前节点为叶节点,返回当前层数即可。
代 码 C++:
/**
* 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;
if(root->left==NULL && root->right==NULL) return 1;
queue<TreeNode*> q;
q.push(root);
int size;
int level = 0;
while(!q.empty()){
size = q.size();
level++;
while(size>0){
TreeNode* node = q.front();
q.pop();
if(node->left==NULL && node->right==NULL)
return level;
if(node->left!=NULL){
q.push(node->left);
}
if(node->right!=NULL){
q.push(node->right);
}
size--;
}
}
return level;
}
};
运行结果:
执行用时 : 20 ms 内存消耗 : 19.7 MB