给定一个二叉树，返回所有从根节点到叶子节点的路径。

说明: 叶子节点是指没有子节点的节点。

示例:

输入:

   1
 /   \
2     3
 \
  5

输出: ["1->2->5", "1->3"]

解释: 所有根节点到叶子节点的路径为: 1->2->5, 1->3

Solution

参考 代码随想录 ，分析回溯过程

• 递归法，前序遍历

class Solution {
private:

    void traversal(TreeNode* cur, vector<int>& path, vector<string>& result) {
        path.push_back(cur->val);
        // 这才到了叶子节点
        if (cur->left == NULL && cur->right == NULL) {
            string sPath;
            for (int i = 0; i < path.size() - 1; i++) {
                sPath += to_string(path[i]);
                sPath += "->";
            }
            sPath += to_string(path[path.size() - 1]);
            result.push_back(sPath);
            return;
        }
        if (cur->left) {
            traversal(cur->left, path, result);
            path.pop_back(); // 回溯
        }
        if (cur->right) {
            traversal(cur->right, path, result);
            path.pop_back(); // 回溯
        }
    }

public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        vector<int> path;
        if (root == NULL) return result;
        traversal(root, path, result);
        return result;
    }
};

• 代码精简

// @lc code=start
/**
 * 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> res;
        if(root==nullptr) return res;
        if(root->left==nullptr && root->right==nullptr){
            res.push_back(to_string(root->val));
            return res;
        }
        vector<string> leftPaths = binaryTreePaths(root->left);
        for(int i=0; i<leftPaths.size(); ++i){
            res.push_back(to_string(root->val)+"->"+leftPaths[i]);
        }
        vector<string> rightPaths = binaryTreePaths(root->right);
        for(int i=0; i<rightPaths.size(); ++i){
            res.push_back(to_string(root->val)+"->"+rightPaths[i]);
        }
        return res;
    }
};
// @lc code=end

• 迭代法，广度优先搜索 参考**@LeetCode官方**
• 维护两个队列，存储节点和根到该节点的路径。

class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> res;
        if(root==nullptr) return res;
        queue<TreeNode *> qNode;
        queue<string> qPath;

        qNode.push(root);
        qPath.push(to_string(root->val));

        while(!qNode.empty()){
            TreeNode* node = qNode.front();
            string path = qPath.front();
            qNode.pop(); qPath.pop();

            if(node->left==nullptr && node->right==nullptr){
                res.push_back(path);
            } else {
                if(node->left){
                    qNode.push(node->left);
                    qPath.push(path+"->"+to_string(node->left->val));
                }
                if(node->right){
                    qNode.push(node->right);
                    qPath.push(path+"->"+to_string(node->right->val));
                }
            }
        }

        return res;
    }
};

​

java

• 递归法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 * int val;
 * TreeNode left;
 * TreeNode right;
 * TreeNode() {}
 * TreeNode(int val) { this.val = val; }
 * TreeNode(int val, TreeNode left, TreeNode right) {
 * this.val = val;
 * this.left = left;
 * this.right = right;
 * }
 * }
 */
class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        List<Integer> path = new ArrayList<>();
        traversal(root, path, result);
        return result;
    }

    private void traversal(TreeNode cur, List<Integer> path, List<String> result) {
        path.add(cur.val);
        if (cur.left == null && cur.right == null) {
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < path.size() - 1; ++i) {
                sb.append(path.get(i)).append("->");
            }
            sb.append(path.get(path.size() - 1));
            result.add(sb.toString());
        }

        if (cur.left != null) {
            traversal(cur.left, path, result);
            path.remove(path.size() - 1);
        }
        if (cur.right != null) {
            traversal(cur.right, path, result);
            path.remove(path.size() - 1);
        }
    }
}

• 迭代法，层序遍历

class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<>();
        if (root == null) {
            return result;
        }
        Queue<TreeNode> nodeQueue = new LinkedList<>();
        Queue<String> pathQueue = new LinkedList<>();
        nodeQueue.offer(root);
        pathQueue.offer(Integer.toString(root.val));

        while (!nodeQueue.isEmpty()) {
            TreeNode node = nodeQueue.poll();
            String path = pathQueue.poll();
            if (node.left == null && node.right == null) {
                result.add(path);
            } else {
                if (node.left != null) {
                    nodeQueue.offer(node.left);
                    pathQueue.offer(new StringBuffer(path).append("->")
                            .append(node.left.val).toString());
                }
                if (node.right != null) {
                    nodeQueue.offer(node.right);
                    pathQueue.offer(new StringBuffer(path).append("->")
                            .append(node.right.val).toString());
                }
            }
        }
        return result;
    }
}