563. Binary Tree Tilt

最新推荐文章于 2021-02-16 03:26:06 发布

原创最新推荐文章于 2021-02-16 03:26:06 发布 · 193 阅读

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本文介绍了一种计算二叉树每个节点的倾斜值并求得整棵树倾斜总和的方法。倾斜值定义为左子树节点值之和与右子树节点值之和的绝对差值。文章提供了两种实现思路，一种利用递归后序遍历计算，另一种通过单独的辅助函数计算子树的节点值之和。

题目描述：

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.

class TreeNode {
      int val;
      TreeNode left;
      TreeNode right;
      TreeNode(int x) { val = x; }
  }

思路一：

class Solution {
    int tilt = 0;
    public int findTilt(TreeNode root) {
        postOrder(root);
        return tilt;
    }
    public int postOrder(TreeNode root)
    {
        if (root == null)
            return 0;
        int left = postOrder(root.left);
        int right = postOrder(root.right);
        tilt += Math.abs(left - right);
        return left + right + root.val;
    }
}

class Solution {
    public int findTilt(TreeNode root) {
        int[] tilt = new int[1];
        postOrder(root, tilt);
        return tilt[0];
    }
    public int postOrder(TreeNode root, int[] tilt)
    {
        if (root == null)
            return 0;
        int left = postOrder(root.left, tilt);
        int right = postOrder(root.right, tilt);
        tilt[0] += Math.abs(left - right);
        return left + right + root.val;
    }
}

思路二：

class Solution {
    public int findTilt(TreeNode root) {
        if (root == null)
            return 0;
        int curVal = 0;
        curVal = Math.abs(SumSubTree(root.left) - SumSubTree(root.right));
        return curVal + findTilt(root.left) + findTilt(root.right);
    }
    public int SumSubTree(TreeNode root)
    {
        if (root == null)
            return 0;
        return root.val + SumSubTree(root.left) + SumSubTree(root.right);
    }
}