leetcode-257-Binary-Tree-Paths

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描述

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

For example, given the following binary tree: 

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5
   1
 /   \
2     3
 \
  5

All root-to-leaf paths are: 

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

分析

树的深度优先遍历。

解决方案1(C++)

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    vector<string> binaryTreePaths(TreeNode* root) {
        vector<string> result;
        if(root == NULL) {
            return result;
        }
        path(root, result, to_string(root->val));
        return result;
    }
    
    void path(TreeNode* root, vector<string>& result, string str_buff) {
        if(root->left==NULL && root->right==NULL) {
            result.push_back(str_buff);
        }
        if(root->left != NULL) {
            path(root->left, result, str_buff+"->"+to_string(root->left->val));
        }
        if(root->right != NULL) {
            path(root->right, result, str_buff+"->"+to_string(root->right->val));
        }
    }
};

解决方案2(Java)

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<String> binaryTreePaths(TreeNode root) {
        List<String> result = new ArrayList<>();
        if(root == null) {
            return result;
        }
        String str_buff = new String();
        path(root, result, (root.val)+"");
        return result;
    }
    
    private void path(TreeNode root, List<String> result, String str_buff) {
        if(root.left == null && root.right == null) {
            result.add(str_buff.toString());
        }
        
        if(root.left != null) {
            path(root.left, result, str_buff+"->"+(root.left.val));
        }
        if(root.right != null) {
            path(root.right, result, str_buff+"->"+(root.right.val));
        }
    }
}

解决方案3(Python)

相关问题

(M) Path Sum II

题目来源