12
05/2015
[LeetCode] Minimum Depth of Binary Tree
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.
解题思路：
这里的深度表示叶子节点到根节点的距离。比如[1, 2]的最小深度为2，而不是1。其代码如下：
/**
* 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;
}else if(root->left==NULL&&root->right==NULL){
return 1; //表示root是叶子节点
}else if(root->left!=NULL&&root->right==NULL){
return 1 + minDepth(root->left);
}else if(root->left==NULL&&root->right!=NULL){
return 1 + minDepth(root->right);
}else{
return 1 + min(minDepth(root->left), minDepth(root->right));
}
}
};二次刷题（2015-08-04）
这道题一定要弄清题意，这里的高度只根节点到叶子节点的路径长度。
/**
* 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;
}
if(root->left!=NULL && root->right ==NULL){
return 1 + minDepth(root->left);
}
if(root->left==NULL && root->right != NULL){
return 1 + minDepth(root->right);
}
return 1 + min(minDepth(root->left), minDepth(root->right));
}
};
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