leetcode 563. Binary Tree Tilt

Given a binary tree, return the tilt of the whole tree.

The tilt of a tree node is defined as the absolute difference between the sum of all left subtree node values and the sum of all right subtree node values. Null node has tilt 0.

The tilt of the whole tree is defined as the sum of all nodes' tilt.

Example:

Input: 
         1
       /   \
      2     3
Output: 1
Explanation: 
Tilt of node 2 : 0
Tilt of node 3 : 0
Tilt of node 1 : |2-3| = 1
Tilt of binary tree : 0 + 0 + 1 = 1

Note:

1. The sum of node values in any subtree won't exceed the range of 32-bit integer.
2. All the tilt values won't exceed the range of 32-bit integer.

 

感觉将先序遍历，后序遍历都折腾了。注意nonlocal

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

class Solution(object):
    def findTilt(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.build_sum(root)
        self.build_tilt(root)
        sum_val = 0
        def dfs(node):
            if not node: return 0
            nonlocal sum_val
            sum_val += node.val
            dfs(node.left)
            dfs(node.right)        
        dfs(root)
        return sum_val
    
    def build_sum(self, node):
        if not node: return 0
        l_val = self.build_sum(node.left)
        r_val = self.build_sum(node.right)
        node.val += l_val+r_val
        return node.val
    
    def build_tilt(self, node):
        if not node: return 0
        node.val = abs((node.left and node.left.val or 0) - (node.right and node.right.val or 0))
        self.build_tilt(node.left)
        self.build_tilt(node.right)

一次性求解的：

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

class Solution(object):
    def findTilt(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.sum_val = 0
        self.build_sum(root)
        return self.sum_val
        
    
    def build_sum(self, node):
        if not node: return 0, 0
        l_sum_val, l_tilt_val  = self.build_sum(node.left)
        r_sum_val, r_tilt_val = self.build_sum(node.right)        
        tilt_val = abs(l_sum_val-r_sum_val) 
        self.sum_val += tilt_val
        return (node.val+l_sum_val+r_sum_val, tilt_val)            

再度精简：

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

class Solution(object):
    def findTilt(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.sum_val = 0
        self.build_sum(root)
        return self.sum_val
            
    def build_sum(self, node):
        if not node: return 0
        l_sum_val  = self.build_sum(node.left)
        r_sum_val = self.build_sum(node.right)        
        self.sum_val += abs(l_sum_val-r_sum_val) 
        return node.val+l_sum_val+r_sum_val           

后序遍历的迭代解法不写了，还不能一次性写对。。。