[LeetCode] 129. Sum Root to Leaf Numbers 求根到叶节点数字之和

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.

给一个只含有数字0-9的二叉树，每一个从根节点到叶节点的路径代表一个数字，求所有这些数字的和。

解法1：递归，累加所有路径上节点的值，每多一层之前的值要扩大十倍。

解法2: 迭代，while循环，用stack或queue来存下一层的节点和之前的和。TLE

Java:

public int sumNumbers(TreeNode root) {
	return sum(root, 0);
}

public int sum(TreeNode n, int s){
	if (n == null) return 0;
	if (n.right == null && n.left == null) return s*10 + n.val;
	return sum(n.left, s*10 + n.val) + sum(n.right, s*10 + n.val);
}

Python:

# Time:  O(n)
# Space: O(h), h is height of binary tree
class TreeNode(object):
    def __init__(self, x):
        self.val = x
        self.left = None
        self.right = None

class Solution(object):
    # @param root, a tree node
    # @return an integer
    def sumNumbers(self, root):
        return self.sumNumbersRecu(root, 0)

    def sumNumbersRecu(self, root, num):
        if root is None:
            return 0

        if root.left is None and root.right is None:
            return num * 10 + root.val

        return self.sumNumbersRecu(root.left, num * 10 + root.val) + self.sumNumbersRecu(root.right, num * 10 + root.val)　

Python:  bfs + stack

def sumNumbers1(self, root):
    if not root:
        return 0
    stack, res = [(root, root.val)], 0
    while stack:
        node, value = stack.pop()
        if node:
            if not node.left and not node.right:
                res += value
            if node.right:
                stack.append((node.right, value*10+node.right.val))
            if node.left:
                stack.append((node.left, value*10+node.left.val))
    return res

Python:  bfs + queue

# 
def sumNumbers2(self, root):
    if not root:
        return 0
    queue, res = collections.deque([(root, root.val)]), 0
    while queue:
        node, value = queue.popleft()
        if node:
            if not node.left and not node.right:
                res += value
            if node.left:
                queue.append((node.left, value*10+node.left.val))
            if node.right:
                queue.append((node.right, value*10+node.right.val))
    return res

Python:  Recursive

def sumNumbers(self, root):
    self.res = 0
    self.dfs(root, 0)
    return self.res
    
def dfs(self, root, value):
    if root:
        #if not root.left and not root.right:
        #    self.res += value*10 + root.val
        self.dfs(root.left, value*10+root.val)
        #if not root.left and not root.right:
        #    self.res += value*10 + root.val
        self.dfs(root.right, value*10+root.val)
        if not root.left and not root.right:
            self.res += value*10 + root.val　

C++:

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    int sumNumbers(TreeNode *root) {
        return sumNumbersDFS(root, 0);
    }
    int sumNumbersDFS(TreeNode *root, int sum) {
        if (!root) return 0;
        sum = sum * 10 + root->val;
        if (!root->left && !root->right) return sum;
        return sumNumbersDFS(root->left, sum) + sumNumbersDFS(root->right, sum);
    }
};

　　

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