LeetCode 563. Binary Tree Tilt (C++)

最新推荐文章于 2021-01-24 22:01:06 发布

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#C++ #LeetCode #二叉树

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本文介绍了一种计算二叉树整体倾斜的方法。首先定义了节点倾斜的概念，即左子树节点值之和与右子树节点值之和的绝对差。接着介绍了两种实现方式：一种是先计算每个节点的倾斜值再进行累加；另一种是在后序遍历过程中同时计算节点值之和及倾斜值，提高了计算效率。

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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:

The sum of node values in any subtree won't exceed the range of 32-bit integer.
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;
        
    }
};