129. Sum Root to Leaf Numbers.md

129. Sum Root to Leaf Numbers

Question

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.

Note: A leaf is a node with no children.

Example:

Input: [1,2,3]
    1
   / \
  2   3
Output: 25
Explanation:
The root-to-leaf path 1->2 represents the number 12.
The root-to-leaf path 1->3 represents the number 13.
Therefore, sum = 12 + 13 = 25.

Example 2:

Input: [4,9,0,5,1]
    4
   / \
  9   0
 / \
5   1
Output: 1026
Explanation:
The root-to-leaf path 4->9->5 represents the number 495.
The root-to-leaf path 4->9->1 represents the number 491.
The root-to-leaf path 4->0 represents the number 40.
Therefore, sum = 495 + 491 + 40 = 1026.

Thinking:

• Method 1:DFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;
        List<Integer> result = new ArrayList<>();
        dfs(root, 0, result);
        int sum = result.get(0);
        for(int i = 1; i < result.size(); i++)
            sum += result.get(i);
        return sum;
    }
    private void dfs(TreeNode root, int sum, List<Integer> result){
        sum = sum * 10 + root.val;
        if(root.left == null && root.right == null){
            result.add(sum);
            return;
        }
        if(root.left != null)
            dfs(root.left, sum, result);
        if(root.right != null)
            dfs(root.right, sum, result);
    }
}

二刷

典型的dfs，深度优先。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        if(root == null) return 0;
        List<Integer> temp = new LinkedList<>();
        dfs(root, temp, 0);
        int result = 0;
        for(Integer s : temp) result += s;
        return result;
    }
    private void dfs(TreeNode node, List<Integer> sum, int cur){
        cur = cur * 10 + node.val;
        if(node.left == null && node.right == null)
            sum.add(cur);
        else{
            if(node.left != null) dfs(node.left, sum, cur);
            if(node.right != null) dfs(node.right, sum, cur);
        }
    }
}

Third Time

• Method 1: dfs

   /**
    * Definition for a binary tree node.
    * public class TreeNode {
    *     int val;
    *     TreeNode left;
    *     TreeNode right;
    *     TreeNode(int x) { val = x; }
    * }
    */
   class Solution {
   	private List<Integer> path;
   	public int sumNumbers(TreeNode root) {
   		this.path = new ArrayList<>();
   		dfs(root, 0);
   		int res = 0;
   		for(Integer n : path)
   			res += n;
   		return res;
   	}
   	private void dfs(TreeNode node, int cur){
   		if(node == null){
   			this.path.add(cur);
   			return;
   		}
   		cur = cur * 10 + node.val;
   		if(node.left == null && node.right == null) dfs(null, cur);
   		if(node.left != null) dfs(node.left, cur);
   		if(node.right != null) dfs(node.right, cur);
   	}
   }

Fourth Time

• Method 1: dfs

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