129 Title: Roots and leaves to node sum of the numbers

One. Problem Description

Given a binary tree, each node that are stored in a digital 0-9, from the root to each leaf node represents a number of paths.

For example, the path from the root to the leaf node 1-> 2-> 3 represents the number 123.

Calculated from the root to the leaf nodes generated by the sum of all the numbers.

Description: leaf node is a node has no child nodes.

Example 1:

Input: [1,2,3]

    1

   / \

  2   3

Output: 25

Explanation:

The path from the root to the leaf node 1-> 2 represents the number 12.

The path from the root to the leaf node 1-> 3 represents the number 13.

Thus, digital sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]

    4

   / \

  9   0

 / \

5   1

Output: 1026

Explanation:

The path from the root to the leaf node 4-> 9-> 5 represents the number 495.

The path from the root to the leaf node 4-> 9-> 1 represents the number 491.

The path from the root to the leaf node 4-> 0 represents the number 40.

Thus, digital sum = 495 + 491 + 40 = 1026.

two. Problem-solving ideas

Body idea: the method is recursive sequence traversal + solved.

Step a: Construction of a recursive function (root root node represents the current representative number from the root to the current node value)

Step two: the root node first determines whether there is a child node, if the value of the root node number * 10 + new child node into the recursive function until the child node is completed traversed, layer by layer returns all values.

three. Results of the

When execution: 1 ms, beat the 63.67% of all users in java submission

Memory consumption: 34.5 MB, defeated 47.62% of all users in java submission

four. Java code

class Solution {
    public int sumNumbers(TreeNode root) {
          if(root==null) {
                return 0;
            }else {
                return getNumber(root,root.val);
            }
    }
   public int getNumber(TreeNode root,int number) {
        if(root.left==null&&root.right==null) {
            return number;
        }
        int templeft=0;
        int tempright=0;
        if(root.left!=null) {
            templeft=getNumber(root.left,root.left.val+number*10);
        }
        if(root.right!=null) {
            tempright=getNumber(root.right,root.right.val+number*10);
        }
        return templeft+tempright;
        
    } 
}