GuilinDev

Lc0563

05 August 2008

563 Binary Tree Tilt

题目

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:

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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.

分析

从问题的描述，需要在给定树的每个结点处找到其坡度，并将所有的坡度相加以获得最终结果。要找出任意结点的坡度，我们需要求出该结点的左子树上所有结点和以及其右子树上全部结点和的差值。

因此，为了找出解决方案，使用递归函数，在任何结点调用该函数，都会返回当前结点下面（包括其自身）的结点和。借助于任何结点的左右子结点的这一和值，可以直接获得该结点所对应的坡度。

代码

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    int tilt = 0;
    public int findTilt(TreeNode root) {
        traverse(root);
        return tilt;
    }
    private int traverse(TreeNode node) {
        // 后序，直接按照题目给的例子的顺序来写
        if (node == null) {
            return 0;
        }
        int left = traverse(node.left);
        int right = traverse(node.right);
        tilt += Math.abs(left - right); // 每一轮更新坡度
        return left + right + node.val;
    }
}
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