Sum Root to Leaf Numbers

Given a binary tree containing digits from 0-9 only, each root-to-leaf
path could represent a number.

An example is the root-to-leaf path 1->2->3 which represents the
number 123.

Find the total sum of all root-to-leaf numbers.

For example,

1    / \   2   3 The root-to-leaf path 1->2 represents the number 12. The root-to-leaf path 1->3 represents the number 13.

Return the sum = 12 + 13 = 25.

递归解法

思路

递归条件是把当前的sum10并且加上当前节点传入下一函数,进行递归,最终把左右子树的总和相加。结束条件是如果到了叶节点 返回sum 10 + 叶节点的值 如果是空节点直接返回0

复杂度

时间是O(n),空间是栈大小O(logn)

代码

public int sumNumbers(TreeNode root) {
        return helper(root, 0);
    }
public int helper(TreeNode root, int sum) {
        if (root == null) {
            return 0;
        }
        if (root.left == null && root.right == null) {
            return sum * 10 + root.val;
        }
        int left = helper(root.left, sum * 10 + root.val);
        int right = helper(root.right, sum * 10 + root.val);
        return left + right;
    }

lpy1990
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