[python] 2019-03-07

.938. Range Sum of BST (不会)

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938. Range Sum of BST

1) description

Given the root node of a binary search tree, return the sum of values of all nodes with value between L and R (inclusive).

The binary search tree is guaranteed to have unique values.

Example 1:

Input: root = [10,5,15,3,7,null,18], L = 7, R = 15
Output: 32
Example 2:
Input: root = [10,5,15,3,7,13,18,1,null,6], L = 6, R = 10
Output: 23

Note:

The number of nodes in the tree is at most 10000.
The final answer is guaranteed to be less than 2^31.

2) solution---other's

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

class Solution:
    def rangeSumBST(self, root: TreeNode, L: int, R: int) -> int:
        def dfs(node):
            if node:
                if L <= node.val <= R:
                    self.ans += node.val
                if L < node.val:
                    dfs(node.left)
                if node.val < R:
                    dfs(node.right)

        self.ans = 0
        dfs(root)
        return self.ans

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

class Solution:
    def rangeSumBST(self, root: TreeNode, L: int, R: int) -> int:
        # dfs 
        # add to output if node.val is within range
        
        stack = [root]
        out = 0
        
        while len(stack) > 0:
            node = stack.pop()
            
            if node.val >= L and node.val <= R:
                out += node.val
                
            if node.left and node.val > L:
                stack.append(node.left)
            if node.right and node.val < R:
                stack.append(node.right)
        return out