129. Sum-Root-to-Leaf-Numbers

difficulty: Medium

section pre{ background-color: #eee; border: 1px solid #ddd; padding:10px; border-radius: 5px; }

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.

Method One

可以归到112 path sum 一类

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int sumNumbers(TreeNode root) {
        if( root == null ){
            return 0;
        }
        int sum = 0;
        ArrayDeque<TreeNode> stack = new ArrayDeque<>();
        ArrayDeque<Integer> vals = new ArrayDeque<>();
        
        stack.push(root);
        vals.push(root.val);
        while(!stack.isEmpty()){
            TreeNode curNode = stack.pop();
            int curVal = vals.pop();
            
            if( curNode.left != null ){
                stack.push(curNode.left);
                vals.push(curVal*10 + curNode.left.val);
            }
            if( curNode.right !=  null ){
                stack.push(curNode.right);
                vals.push(curVal*10 + curNode.right.val);
            }
            
            if( curNode.left == null && curNode.right == null){
                sum += curVal;
            }
        }
        return sum;
    }
}
​