《leetcode-php》二叉树从根到叶组成的数求和

本文链接：https://blog.youkuaiyun.com/less_cold/article/details/98880789

本文探讨了如何在二叉树中寻找所有从根节点到叶子节点的路径，并计算这些路径所代表数字的总和。通过递归算法，每向下一层，路径数更新为上一层的10倍加上当前节点值，最终返回左子树和右子树路径和的总和。

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Given a binary tree containing digits from0-9only, each root-to-leaf path could represent a number.
An example is the root-to-leaf path1->2->3which represents the number123.
Find the total sum of all root-to-leaf numbers.
For example,
1
/ \
2 3

The root-to-leaf path1->2represents the number12.
The root-to-leaf path1->3represents the number13.
Return the sum = 12 + 13 =25.
从根到叶组成所有数求和。

思路：
每向下一层，都是上面的和*10+当前值作为组成的数。

分叉的情况就是左边+右边

<?php
class TreeNode {
    public $val;
    public $left  = null;
    public $right = null;
    public function __construct($val) {
        $this->val = $val;
    }
}
function sumNumbers($head, $ret) {
    if ($head->left == null && $head->right == null) {
        return $ret * 10 + $head->val;
    }
    $leftRet  = 0;
    $rightRet = 0;
    if ($head->left !== null) {
        $leftRet = sumNumbers($head->left, $ret * 10 + $head->val);
    }
    if ($head->right !== null) {
        $rightRet = sumNumbers($head->right, $ret * 10 + $head->val);
    }
    return $leftRet + $rightRet;
}
$node1 = new TreeNode(1);
$node2 = new TreeNode(2);
$node3 = new TreeNode(3);
$node4 = new TreeNode(4);
$node1->left  = $node2;
$node1->right = $node3;
$node3->right = $node4;
$ret = sumNumbers($node1, 0);
print $ret;