[LeetCode] Binary Tree Tilt

标签：abs   斜度   log   计算   solution   ati   tween   block   div   

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.

 使用后序遍历递归计算每一个节点的倾斜度，维护一个节点与子树和作为返回值返回上层递归调用。最后可得整棵树的倾斜度。

class Solution {
public:
    int res = 0;
    int findTilt(TreeNode* root) {
        dfs(root);
        return res;
    }
    int dfs(TreeNode* root) {
        if (root == nullptr)
            return 0;
        int left = dfs(root->left);
        int right = dfs(root->right);
        res += abs(left - right);
        return left + right + root->val;
    }
};
// 22 ms

[LeetCode] Binary Tree Tilt

标签：abs   斜度   log   计算   solution   ati   tween   block   div   

原文地址：http://www.cnblogs.com/immjc/p/7183496.html