Leetcode 988. Smallest String Starting From Leaf (二叉树遍历好题)

  1. Smallest String Starting From Leaf
    Medium
    1.6K
    227
    Companies
    You are given the root of a binary tree where each node has a value in the range [0, 25] representing the letters ‘a’ to ‘z’.

Return 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: root = [0,1,2,3,4,3,4]
Output: “dba”
Example 2:

Input: root = [25,1,3,1,3,0,2]
Output: “adz”
Example 3:

Input: root = [2,2,1,null,1,0,null,0]
Output: “abc”

Constraints:

The number of nodes in the tree is in the range [1, 8500].
0 <= Node.val <= 25

解法1:遍历二叉树即可。
注意这里是要找到顺序最小的字符串,所以gMaxPath要初始化成ascii很大的字符串。而ascii table里面,数字<大写字母<小写字母。在前128个ascii code里面,比’z’还大的只有’{‘,’}‘,‘|’和DEL。所以可以初始化成"{}"。

/**
 * 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) {
        helper(root, "");
        return gMaxPath;
    }
private:
    string gMaxPath = "{}";//'{' or '}' > 'z'
    void helper(TreeNode *root, string path) {
        if (!root) return;
        path += 'a' + root->val;
        if (!root->left && !root->right) {
            reverse(path.begin(), path.end());
            gMaxPath = min(gMaxPath, path);
            return;
        }
        helper(root->left, path);
        helper(root->right, path);
        return;
    }
};

二刷:path改成引用。
需要注意:

  1. 如果不是引用,可以直接调用helper(root, “”),但是如果是引用的话,需要先定义string path = “”, 然后调用helper(root, path)。
  2. 在 if (!root->left && !root->right) {}里面return前也要 path.pop_back();
/**
 * 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) {
        string path = "";
        helper(root, path);
        return gMaxPath;
    }
private:
    string gMaxPath = "{}";//'{' or '}' > 'z'
    void helper(TreeNode *root, string &path) {
        if (!root) return;
        path += 'a' + root->val;
        if (!root->left && !root->right) {
            string tmp = path;
            reverse(tmp.begin(), tmp.end());
            gMaxPath = min(gMaxPath, tmp);
            path.pop_back();
            return;
        }
        helper(root->left, path);
        helper(root->right, path);
        path.pop_back();
        return;
    }
};

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