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2014
11-19

LeetCode-Sum Root to Leaf Numbers[二叉树]

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.

标签: Tree Depth-first Search
分析

代码1

// LeetCode, Decode Ways
// 时间复杂度O(n),空间复杂度O(logn)
class Solution {
public:
    int sumNumbers(TreeNode *root) {
        return dfs(root, 0);
    }
private:
    int dfs(TreeNode *root, int sum) {
        if (root == nullptr) return 0;
        if (root->left == nullptr && root->right == nullptr)
            return sum * 10 + root->val;

        return dfs(root->left, sum * 10 + root->val) +
                dfs(root->right, sum * 10 + root->val);
    }
};

Java代码:

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    
    def sumNumRecr(self,root, parent):
        if root.left == None and root.right == None:
            self.sum +=(parent*10 + root.val)
            
        if root.left != None:
            self.sumNumRecr(root.left, parent*10 + root.val)
        
        if root.right != None:
            self.sumNumRecr( root.right, parent*10 + root.val)
            
            
    # @param root, a tree node
    # @return an integer
    def sumNumbers(self, root):
        if(root == None):
            return 0
        self.sum = 0 
        self.sumNumRecr(root,0)
        return self.sum
        
    
        
        

  1. 第二个方法挺不错。NewHead代表新的头节点,通过递归找到最后一个节点之后,就把这个节点赋给NewHead,然后一直返回返回,中途这个值是没有变化的,一边返回一边把相应的指针方向颠倒,最后结束时返回新的头节点到主函数。