问题:给定二叉树,找其最小深度(叶子节点到根节点最短路径上的节点个数)
输入:TreeNode* root
输出:minimum_depth
思路:递归。一开始想着,只要一个儿子结点为NULL,便得到了最小深度,其实不是的,输入样例[1,2]便可证明。
看了[1]才明白自己错在哪里了,题目要求的是叶子结点到根结点最短路径,必须找到叶子结点,才可退出递归。
/**
* 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 minimum_depth=0;
int minDepth(TreeNode* root) {
if(root==NULL)
return 0;
else if(root->left==NULL)
minimum_depth = 1+minDepth(root->right);
else if(root->right==NULL)
minimum_depth = 1+minDepth(root->left);
else
minimum_depth = min(minDepth(root->left),minDepth(root->right))+1;
return minimum_depth;
}
};
大神用层序遍历(即广度优先遍历)做的,贴代码分享,一起学习!
int minDepth(TreeNode* root) {
if (root == NULL) return 0;
queue<TreeNode*> Q;
Q.push(root);
int i = 0;
while (!Q.empty()) {
i++;
int k = Q.size();
for (int j=0; j<k; j++) {
TreeNode* rt = Q.front();
if (rt->left) Q.push(rt->left);
if (rt->right) Q.push(rt->right);
Q.pop();
if (rt->left==NULL && rt->right==NULL) return i;
}
}
return -1; //For the compiler thing. The code never runs here.
}