二叉树路径 - 我的代码出了什么问题

Question

Here is the binary tree paths problem: Given a binary tree, return all root-to-leaf paths.

For example, given the following binary tree:

   1
 /   \
2     3
 \
  5

All root-to-leaf paths are:

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

And here is my Javascript code:

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {string[]}
 */
var binaryTreePaths = function(root) {
    var paths = [];
    if(!root) return [];
    if(root.left == null && root.right == null){
        if(paths.length == 0) return [""+root.val];
        else return root.val;
    } 
    else{
        if(root.left) paths.push(root.val + "->" + binaryTreePaths(root.left))
        if(root.right) paths.push(root.val + "->" + binaryTreePaths(root.right))
    }

    return paths;
};

Test Case:

Input:

[1,2,3,5,6]

Output:

["1->2->5,2->6","1->3"]

Expected:

["1->2->5","1->2->6","1->3"]

Why is my code's output not returning the full path of "1->2->6" ?

Answer 1

When you make a recursive call, your function will return an array. You can't just push the concatenation of that array with the prefix string; you need to iterate through each of the returned subpaths and build a separate path to push onto the array:

var binaryTreePaths = function(root) {
    var paths = [];
    if(!root) return [];
    if(root.left == null && root.right == null){
        if(paths.length == 0) return [""+root.val];
        else return root.val;
    } 
    else{
        if(root.left) 
          binaryTreePaths(root.left).forEach(function(lp) {
            paths.push(root.val + "->" + lp);
          });
        if(root.right) 
          binaryTreePaths(root.right).forEach(function(rp) {
            paths.push(root.val + "->" + rp);
          });
    }

    return paths;
};