[LeetCode] Binary Tree Tilt

Posted immjc

tags:

篇首语:本文由小常识网(cha138.com)小编为大家整理,主要介绍了[LeetCode] Binary Tree Tilt相关的知识,希望对你有一定的参考价值。

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

 

以上是关于[LeetCode] Binary Tree Tilt的主要内容,如果未能解决你的问题,请参考以下文章

[Leetcode] Binary search tree --Binary Search Tree Iterator

Leetcode[110]-Balanced Binary Tree

[Leetcode] Binary tree -- 501. Find Mode in Binary Search Tree

[Lintcode]95. Validate Binary Search Tree/[Leetcode]98. Validate Binary Search Tree

LeetCode 110. Balanced Binary Tree 递归求解

[Leetcode] Balanced Binary Tree