leetcode 563. Binary Tree Tilt

Given the root of a binary tree, return the sum of every tree node's tilt.

The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values. If a node does not have a left child, then the sum of the left subtree node values is treated as 0. The rule is similar if there the node does not have a right child.

Example 1:

Input: root = [1,2,3]
Output: 1
Explanation: 
Tilt of node 2 : |0-0| = 0 (no children)
Tilt of node 3 : |0-0| = 0 (no children)
Tile of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3)
Sum of every tilt : 0 + 0 + 1 = 1

Example 2:

Input: root = [4,2,9,3,5,null,7]
Output: 15
Explanation: 
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 5 : |0-0| = 0 (no children)
Tilt of node 7 : |0-0| = 0 (no children)
Tilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5)
Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)
Tilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16)
Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15

Example 3:

Input: root = [21,7,14,1,1,2,2,3,3]
Output: 9

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -1000 <= Node.val <= 1000

题目难度：简单题

思路：我们需要返回所有node的tilt，想法就是遍历到某个node时，计算出其tilt，并把它累加到一个变量中，最后返回。计算某个node的tilt时，需要左子树的各个node值之和，以及右子树的各个node值之和，因此新建一个函数，返回值为树的node值之和。进一步理解见代码：

C++代码：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
private:
    int findTilt(TreeNode *root, int &sum) {
        //root为传入树的根节点，sum表示以root为跟节点的所有节点tilt值之和
        if (root == nullptr) { //如果root为空，返回值为0
            return 0;
        }
        int left = findTilt(root->left, sum);
        int right = findTilt(root->right, sum);
        sum += abs(left - right);
        return left + right + root->val;
    }
public:
    int findTilt(TreeNode* root) {
        int sum = 0;
        findTilt(root, sum);
        return sum;
    }
};

时间复杂度：$O(n)$,空间复杂度：$O(n)$

python3代码：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def findTilt(self, root: TreeNode) -> int:
        tTilt = 0
        
        def valueSum(node):
            nonlocal tTilt
            if not node:
                return 0
            
            left = valueSum(node.left)
            right = valueSum(node.right)
            tTilt += abs(left - right)
            
            return left + right + node.val
        
        valueSum(root)
        
        return tTilt