LeetCode-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.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int x) { val = x; }
 * }
 */

递归解法:

public class Solution 
{
    public int SumNumbers(TreeNode root) 
    {
        if(root==null)
            return 0;
        return sumR(root, 0);
    }
    
    public int sumR(TreeNode root, int x)
    {
        if (root.right == null && root.left == null)
            return 10 * x + root.val;
        int val = 0;
        if (root.left != null)
            val += sumR(root.left, 10 * x + root.val);
        if (root.right != null)
            val += sumR(root.right, 10 * x + root.val);
        return val;
    }
}


非递归
 public int SumNumbers(TreeNode root)
            {
                if (root == null)
                    return 0;
                Stack stackNode = new Stack();
                Stack stackNum = new Stack();
                int result = 0;
                stackNode.Push(root);
                stackNum.Push(root.val);
                while (stackNode.Count != 0)
                {
                    TreeNode node = (TreeNode)stackNode.Pop();
                    int val = (int)stackNum.Pop();
                    if (node.left == null && node.right == null)
                    {
                        result += val;
                    }
                    if (node.left != null)
                    {
                        stackNode.Push(node.left);
                        stackNum.Push(val * 10 + node.left.val);
                    }
                    if (node.right != null)
                    {
                        stackNode.Push(node.right);
                        stackNum.Push(val * 10 + node.right.val);
                    }
                }
                return result;
            }



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