LeetCode 563. Binary Tree Tilt (C++)

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     3Output: 1Explanation: Tilt of node 2 : 0Tilt of node 3 : 0Tilt of node 1 : |2-3| = 1Tilt 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.

思路：先求每个node的tilt，再前序遍历相加得到整个tree的tilt。

class Solution {public:    int findTilt(TreeNode* root) {        int tilt = 0;        getTreeTilt(root, tilt);        return tilt;    }        void getTreeTilt(TreeNode* root, int & tilt) {        if (!root) return;        tilt += getNodeTilt(root);        getTreeTilt(root->left, tilt);        getTreeTilt(root->right, tilt);    }        int getNodeTilt(TreeNode* root) {        return abs(getSum(root->left) - getSum(root->right));    }        int getSum(TreeNode* root) {        if (!root) return 0;        return (root->val + getSum(root->left) + getSum(root->right));    }};

显然，这样效率是比较低的。因为getSum是重复计算的。可以用后序遍历求sum的同时，计算出每个node的tilt。代码如下

class Solution {public:    int findTilt(TreeNode* root) {        if(root == NULL) return 0;                int res = 0;                postorder(root, res);                return res;    }private:    int postorder(TreeNode* root, int& res){        if(root == NULL) return 0;                int leftsum= postorder(root->left,res);                int rightsum = postorder(root->right,res);                res += abs(leftsum - rightsum);                return leftsum + rightsum + root->val;            }};