leetcode 563. Binary Tree Tilt

最新推荐文章于 2022-03-25 15:51:31 发布

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本文介绍了一种计算二叉树每个节点的倾斜值的方法，并通过递归遍历实现了整体二叉树的倾斜值求和。文章提供了两种实现方案，一种是通过定义类字段来累积节点的倾斜值，另一种则是采用后序遍历的方法来更新结果。

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

首先必须要有一个子树求和的过程，返回子树的和。然后在每次返回前还需计算题目需要的tilt。那么最简单的当然就是将tilt设为类的字段。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int tilt=0;
    public int findTilt(TreeNode root) {
        int t = childsum(root);

        return tilt;
    }

    int childsum(TreeNode root){
        if(root==null) return 0;
        int left = childsum(root.left);
        int right = childsum(root.right);
        tilt += Math.abs(left-right);
        return left+right+root.val;
    }
}

top answer的叫法是 post-order traversal

public class Solution {
    int result = 0;

    public int findTilt(TreeNode root) {
        postOrder(root);
        return result;
    }

    private int postOrder(TreeNode root) {
        if (root == null) return 0;

        int left = postOrder(root.left);
        int right = postOrder(root.right);

        result += Math.abs(left - right);

        return left + right + root.val;
    }
}