129. 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.

一刷
题解：DFS。leaf的左右子节点均为空。
Time Complexity - O(n)， Space Complexity - O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sumNumbers(TreeNode root) {
        return sum(root, 0);
    }
    
    public int sum(TreeNode n, int s){
        if(n == null) return 0;
        if(n.right == null && n.left == null) return s*10 + n.val;
        return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val);
    }
}

二刷：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int total = 0;
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;
        sum(0, root);
        return total;
    }
    
    public void sum(int prev, TreeNode root){
        if(root == null) return;
        if(root.left == null && root.right == null) {
            total += prev*10 + root.val;
            return;
        }
        sum(prev*10 + root.val, root.left);
        sum(prev*10 + root.val, root.right);
    }
}

三刷
注意要判断叶子，否则[1,2]这种结构会返回1+12

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int sumNumbers(TreeNode root) {
        return sumNumbers(root, 0);
    }
    
    public int sumNumbers(TreeNode root, int sum){
        if(root == null) return 0;
        if(root.left == null && root.right == null) return sum*10 + root.val;
        return sumNumbers(root.left, 10*sum + root.val) + sumNumbers(root.right, 10*sum + root.val);
    }
}