Depth First Search Algorithm to Compute the Smallest String Star

  • 时间：2020-09-09 13:16:32
  • 分类：网络文摘
  • 阅读：123 次

Given the root of a binary tree, each node has a value from 0 to 25 representing the letters ‘a’ to ‘z’: a value of 0 represents ‘a’, a value of 1 represents ‘b’, and so on.

Find the lexicographically smallest string that starts at a leaf of this tree and ends at the root.

(As a reminder, any shorter prefix of a string is lexicographically smaller: for example, “ab” is lexicographically smaller than “aba”. A leaf of a node is a node that has no children.)

Example 1:
Input: [0,1,2,3,4,3,4]
Output: “dba”
binary-tree1 Depth First Search Algorithm to Compute the Smallest String Starting From Leaf algorithms c / c++ DFS recursive

Example 2:
Input: [25,1,3,1,3,0,2]
Output: “adz”
binary-tree2 Depth First Search Algorithm to Compute the Smallest String Starting From Leaf algorithms c / c++ DFS recursive

Example 3:
Input: [2,2,1,null,1,0,null,0]
Output: “abc”
binary-tree3 Depth First Search Algorithm to Compute the Smallest String Starting From Leaf algorithms c / c++ DFS recursive

Note:
The number of nodes in the given tree will be between 1 and 8500.
Each node in the tree will have a value between 0 and 25.

Recursive Depth First Search to Compute the Smallest String Starting From Leaf

We can start from the Root, perform a recursive DFS (Depth First Search) algorithm, and pass the string along down to the leaves. When we reach a leaf node, we construct the final string, and store the smallest one (alphabetically).

All nodes are visisted exactly once. When we finish searching the tree, we have the answer. The time complexity is O(N). In this particular case, the DFS is also the bruteforce algorithm.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string smallestFromLeaf(TreeNode* root) {
        dfs(root, "");
        return ans;
    }
private:
    string ans = "";
    void dfs(TreeNode* root, string cur) {
        if (!root) return;
        if ((root->left == nullptr) && (root->right == nullptr)) {
            cur = (char)(97 + root->val) + cur;
            if ((ans == "") || (cur < ans)) {
               ans = cur; 
            }
            return;
        }
        dfs(root->left, (char)(97 + root->val) + cur);
        dfs(root->right,(char)(97 + root->val) + cur);
    }
};
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string smallestFromLeaf(TreeNode* root) {
        dfs(root, "");
        return ans;
    }
private:
    string ans = "";
    void dfs(TreeNode* root, string cur) {
        if (!root) return;
        if ((root->left == nullptr) && (root->right == nullptr)) {
            cur = (char)(97 + root->val) + cur;
            if ((ans == "") || (cur < ans)) {
               ans = cur; 
            }
            return;
        }
        dfs(root->left, (char)(97 + root->val) + cur);
        dfs(root->right,(char)(97 + root->val) + cur);
    }
};

Depth First Searching Algorithm using Stack

The Recursion uses implicit stacks. The compiler generates the calling stacks. The recursion code usually is more clean and concise. We can also use a stack to manually emulate the recursive calls.

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string smallestFromLeaf(TreeNode* root) {
        if (!root) return "";
        stack<pair<TreeNode*, string>> st;
        st.push({root, ""});
        string ans = "";
        while (!st.empty()) {
            auto p = st.top();
            st.pop();
            if (p.first->right) {
                st.push({p.first->right, (char)(97 + p.first->val) + p.second});
            }            
            if (p.first->left) {
                st.push({p.first->left, (char)(97 + p.first->val) + p.second});
            }
            if ((p.first->left == nullptr) && (p.first->right == nullptr)) {
                string v = (char)(97 + p.first->val) + p.second;
                if ((ans == "") || (v < ans)) {
                    ans = v;   
                }                
            }
        }
        return ans;
    }
};
/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    string smallestFromLeaf(TreeNode* root) {
        if (!root) return "";
        stack<pair<TreeNode*, string>> st;
        st.push({root, ""});
        string ans = "";
        while (!st.empty()) {
            auto p = st.top();
            st.pop();
            if (p.first->right) {
                st.push({p.first->right, (char)(97 + p.first->val) + p.second});
            }            
            if (p.first->left) {
                st.push({p.first->left, (char)(97 + p.first->val) + p.second});
            }
            if ((p.first->left == nullptr) && (p.first->right == nullptr)) {
                string v = (char)(97 + p.first->val) + p.second;
                if ((ans == "") || (v < ans)) {
                    ans = v;   
                }                
            }
        }
        return ans;
    }
};

During the code interviews, if you are asked this question, it is better to go with the Recursion. But you can follow up with the non-recursive/iterative approach.

–EOF (The Ultimate Computing & Technology Blog) —

推荐阅读：
男人多吃一些香蕉对身体大有好处　 冬至时节常吃6种传统食物可补阳防寒　 南方黑芝麻糊大肠菌群超标 5批次产品不合格　 烹饪技巧之烹调鸡蛋过程中的常见错误　 全新认识冬季当家菜大白菜的营养价值　 那些被人们误传的补充某种营养的食物　 阿胶产品乱象：原料中被混入骡皮马皮　 时令水果橘子橙子柚子营养价值大比拼　 中医食疗：滋阴固肾的6款经典补粥　 美味鸭肉的保健功效与养生食用方法　
评论列表
添加评论