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"]

Example 2:

Input: root = [1]
Output: ["1"]

 

Constraints:

• The number of nodes in the tree is in the range [1, 100].
• -100 <= Node.val <= 100

A:  

递归，同时传入路径作为String参数

/**
 * 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:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> V;
        helper(root, V, to_string(root->val));
        return V;
    }
private:
    void helper(TreeNode* root, vector<string>& V,
                string pre) { // root is not null
        if (!root)
            return;
        if (!root->left && !root->right) {
            V.push_back(pre);
        }
        if (root->left) {
            helper(root->left, V, pre + "->" + to_string(root->left->val));
        }
        if (root->right) {
            helper(root->right, V, pre + "->" + to_string(root->right->val));
        }
    }
};

*************2nd pass**********没有用helper Function*********

/**

* 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:

vector<string> binaryTreePaths(TreeNode* root) {// root is not NULL

vector<string> V;

if(!root->left && ! root->right){

V.push_back(to_string(root->val));

}

if(root->left){

auto leftRes = binaryTreePaths(root->left);

for(auto s: leftRes){

V.push_back(to_string(root->val) + "->" + s);

}

}

if(root->right){

auto rightRes = binaryTreePaths(root->right);

for(auto s: rightRes){

V.push_back(to_string(root->val) + "->" + s);

}

}

return V;

}

};

********Iteratively,  using a stack ****************
但是这样，需要考虑该节点只有一个child 都情况，因此需要判断其有几个child。