[Leetcode][Tree][Depth of Binary Tree]

计算树的深度

1、minimum depth of binary tree

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int minDepth(TreeNode *root) {
13         if (root == NULL) {
14             return 0;
15         }
16         if (root->left == NULL && root->right == NULL) {
17             return 1;
18         }
19         int left = INT_MAX;
20         int right = INT_MAX;
21         if (root->left != NULL) {
22             left = minDepth(root->left);
23         }
24         if (root->right != NULL) {
25             right = minDepth(root->right);
26         }
27         return min(left, right) + 1;
28     }
29 };

总结：刚开始用非常简单的递归，发现wa了，是由于root结点在有一个子树的时候，不算是叶子，需要特殊处理。

又写了一个迭代的版本，也可以用作树的层次遍历。回忆了一下stack和queue的用法。

 1 /**
 2  * Definition for binary tree
 3  * struct TreeNode {
 4  *     int val;
 5  *     TreeNode *left;
 6  *     TreeNode *right;
 7  *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 8  * };
 9  */
10 class Solution {
11 public:
12     int minDepth(TreeNode *root) {
13         int result = INT_MAX;
14         queue<pair<TreeNode*, int>> q;
15         if (root != NULL) {
16             q.push(make_pair(root,1));
17         } else {
18             return 0;
19         }
20         while (!q.empty()) {
21             auto node = q.front().first;
22             auto depth = q.front().second;
23             q.pop();
24             if (node->left == NULL && node->right == NULL) {
25                 result = min(result, depth);
26             }
27             if (node->left != NULL) {
28                 q.push(make_pair(node->left, depth + 1));
29             }
30             if (node->right != NULL) {
31                 q.push(make_pair(node->right, depth + 1));
32             }
33         }
34         return result;
35     }
36 };

2、Maximum depth of binary tree

/**
 * 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 maxDepth(TreeNode *root) {
        if (root == NULL) {
            return 0;
        }
        return max(maxDepth(root->left), maxDepth(root->right)) + 1;
        /*
        if (root->left == NULL && root->right == NULL) {
            return 1;
        }
        int left = INT_MIN;
        int right = INT_MIN;
        if (root->left != NULL) {
            left = maxDepth(root->left);
        }
        if (root->right != NULL) {
            right = maxDepth(root->right);
        }
        return max(left, right) + 1;
        */
    }
};

CE了一次。。又不小心把单词拼错了。。

总结：maximum depth用了非常简单的递归，但是这种方法用在Minimum上就不行，需要对root结点进行特殊的处理。

[Leetcode][Tree][Depth of Binary Tree]