257 - Binary Tree Paths && 112 - Path Sum

257 - Binary Tree Paths

#easy

Given the root of a binary tree, return all root-to-leaf paths in any order.

A leaf is a node with no children.

Example 1:

Input: root = [1,2,3,null,5]
Output: ["1->2->5","1->3"]

Solution 1

/**
 * 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:
    void traversal( TreeNode * root, vector<int> & path, vector<string> & result ) {
	    // 中節點
        path.push_back( root -> val );
        if ( root -> left == NULL && root -> right == NULL ) {
            string one_path = "";
            for ( int i = 0; i < path.size(); i++ ) {
                if ( i == 0 ) one_path = to_string(path[i]);
                else one_path += "->" + to_string(path[i]); 
            }

            result.push_back(one_path);
            return; // 回朔!
        }

		// 左節點
        if ( root -> left ) {
            traversal( root -> left, path, result );
            path.pop_back();
        }

		// 右節點
        if ( root -> right ) {
            traversal( root -> right, path, result );
            path.pop_back();
        }
    }

    vector<string> binaryTreePaths(TreeNode* root) {
        vector<int> path;
        vector<string> result;
        traversal(root, path, result);
        return result;
    }
};

Solution 2

class Solution {
public:
    void traversal( TreeNode * root, vector<int> & path, vector<string> & result ) {
        path.push_back( root -> val );
        if ( root -> left == NULL && root -> right == NULL ) {
            string one_path = "";
            for ( int i = 0; i < path.size(); i++ ) {
                if ( i == 0 ) one_path = to_string(path[i]);
                else one_path += "->" + to_string(path[i]); 
            }

            result.push_back(one_path);
            return;
        }

        if ( root -> left ) {
            traversal( root -> left, path, result );
            path.pop_back();
        }

        if ( root -> right ) {
            traversal( root -> right, path, result );
            path.pop_back();
        }
    }

    vector<string> binaryTreePaths(TreeNode* root) {
        vector<int> path;
        vector<string> result;
        traversal(root, path, result);
        return result;
    }
};

如上代碼精簡了不少，也隱藏了不少東西。

注意在函數定義的時候void traversal(TreeNode* cur, string path, vector<string>& result) ，定義的是string path，每次都是複製賦值，不用使用引用，否則就無法做到回溯的效果。（這裡涉及到C++語法知識）

那麼在如上代碼中，貌似沒有看到回溯的邏輯，其實不然，回溯就隱藏在traversal(cur->left, path + "->", result);中的 path + "->"。 每次函數調用完，path依然是沒有加上"->" 的，這就是回溯了。

112 - Path Sum

#easy

Given the root of a binary tree and an integer targetSum, return true if the tree has a root-to-leaf path such that adding up all the values along the path equals targetSum.

A leaf is a node with no children.

Example 1:

Input: root = [5,4,8,11,null,13,4,7,2,null,null,null,1], targetSum = 22
Output: true
Explanation: The root-to-leaf path with the target sum is shown.

Example 2:

Input: root = [1,2,3], targetSum = 5
Output: false
Explanation: There two root-to-leaf paths in the tree:
(1 --> 2): The sum is 3.
(1 --> 3): The sum is 4.
There is no root-to-leaf path with sum = 5.

Example 3:
Input: root = [], targetSum = 0
Output: false
Explanation: Since the tree is empty, there are no root-to-leaf paths.

Solution

class Solution {
public:
    void traversal( TreeNode * root, int & path_nums, int targetSum, bool & isFound ) {
        if ( root == NULL || isFound ) return;
        path_nums += root -> val;
        if ( root -> left == NULL && root -> right == NULL ) {
            if ( path_nums == targetSum ) isFound = true;
            path_nums -= root -> val;
            return;
        }

        if ( root -> left ) {
            traversal( root -> left, path_nums, targetSum, isFound );
        }

        if ( root -> right ) {
            traversal( root -> right, path_nums, targetSum, isFound );
        }
        
        path_nums -= root -> val;
    }

    bool hasPathSum(TreeNode* root, int targetSum) {
        int path_nums = 0;
        bool isFound = false;
        if ( root == NULL ) return false;
        else {
            traversal( root, path_nums, targetSum, isFound );
            if ( isFound ) return true;
        }

        return false;
    }
};