刷题leetcode--563. Binary Tree Tilt

清疚 2021-10-09 03:50 315阅读 0赞

**2018.1.8路上的风景很精彩！享受过程！![大哭][wail.gif]**

**563. Binary Tree Tilt![大哭][wail.gif]![大哭][wail.gif]![大哭][wail.gif]**

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.

题意：（自己完全理解错了，还是看的大神的。。。）每层都要算倾差值，一层层向上求和后算上一层的差值，最后再把差值想加。

知识点：后序遍历

题意是这样的：

For example:  
      1  
   /      \\  
  2       3  
 /         /  
4       5

Output: 11  
Explanation:  
Tilt of node 4 : 0  
Tilt of node 5 : 0  
Tilt of node 2 : |4-0| = 4  
Tilt of node 3 : |5-0| = 5  
Tilt of node 1 : |sum(2,4)-sum(3,5)| = |6-8| = 2  
Tilt of binary tree : 0 + 0 + 4 + 5 + 2 = 11

public int findTilt(TreeNode root) {
            if (root == null) return 0;
            int curVal = 0;
            curVal = Math.abs(sumSubTree(root.left) - sumSubTree(root.right)); //计算当前层的差值（左子树的和 与 右子树的和 的差）
            return curVal + findTilt(root.left) + findTilt(root.right);  //差值要叠加，包括左右子树的差值。最后返回总差值。
        }
           //子树求和
        private int sumSubTree(TreeNode root) {
            if (root == null) return 0;
            return root.val + sumSubTree(root.left) + sumSubTree(root.right);// 累加求和，当前和是左右子树的和。
        }

[wail.gif]: /images/20211009/66266763c0984528817a3794fbd18b2d.png