257. Binary Tree Paths

• 题目

• https://leetcode.com/problems/binary-tree-paths/discuss/68279/C++-non-recursive-version-and-recursive-version

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

 -C++

/*
follow up: non-recursive version
*/
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if (root == NULL) return res;
        stack<TreeNode*> s;
        stack<string> pathStack;
        s.push(root);
        pathStack.push(to_string(root->val));
        
        while (!s.empty()) {
            TreeNode * curNode = s.top(); s.pop();
            string tmpPath = pathStack.top(); pathStack.pop();
            
            if (curNode->left == NULL && curNode->right == NULL) {
                res.push_back(tmpPath); continue;
            }
            
            if (curNode->left != NULL) {
                s.push(curNode->left);
                pathStack.push(tmpPath + "->" + to_string(curNode->left->val));
            }
            
            if (curNode->right != NULL) {
                s.push(curNode->right);
                pathStack.push(tmpPath + "->" + to_string(curNode->right->val));
            }
        }
        
        return res;
    }
};

//recursive version
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if (root == NULL) return res;
        dfs(root, to_string(root->val), res);
        return res;
    }
    
    void dfs(TreeNode* root, string path, vector<string>& res) {
        if (root->left == NULL && root->right == NULL) {
            res.push_back(path);
        }
        
        if (root->left != NULL)
            dfs(root->left, path + "->" + to_string(root->left->val), res);
        
        if (root->right != NULL)
            dfs(root->right, path + "->" + to_string(root->right->val), res);
    }
};

 -python

# dfs + stack
def binaryTreePaths1(self, root):
    if not root:
        return []
    res, stack = [], [(root, "")]
    while stack:
        node, ls = stack.pop()
        if not node.left and not node.right:
            res.append(ls+str(node.val))
        if node.right:
            stack.append((node.right, ls+str(node.val)+"->"))
        if node.left:
            stack.append((node.left, ls+str(node.val)+"->"))
    return res
    
# bfs + queue
def binaryTreePaths2(self, root):
    if not root:
        return []
    res, queue = [], collections.deque([(root, "")])
    while queue:
        node, ls = queue.popleft()
        if not node.left and not node.right:
            res.append(ls+str(node.val))
        if node.left:
            queue.append((node.left, ls+str(node.val)+"->"))
        if node.right:
            queue.append((node.right, ls+str(node.val)+"->"))
    return res
    
# dfs recursively
def binaryTreePaths(self, root):
    if not root:
        return []
    res = []
    self.dfs(root, "", res)
    return res

def dfs(self, root, ls, res):
    if not root.left and not root.right:
        res.append(ls+str(root.val))
    if root.left:
        self.dfs(root.left, ls+str(root.val)+"->", res)
    if root.right:
        self.dfs(root.right, ls+str(root.val)+"->", res)