Binary Tree Tilt
Given the root of a binary tree, return the sum of every tree node's tilt.
The tilt of a tree node is the absolute difference between the sum of all left subtree node values and all right subtree node values.
If a node does not have a left child, then the sum of the left subtree node values is treated as 0. The rule is similar if the node does not have a right child.
Example 1:
Input: root = [1,2,3]
Output: 1
Explanation:
Tilt of node 2 : |0-0| = 0 (no children)
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 1 : |2-3| = 1 (left subtree is just left child, so sum is 2; right subtree is just right child, so sum is 3)
Sum of every tilt : 0 + 0 + 1 = 1
Example 2:
Input: root = [4,2,9,3,5,null,7]
Output: 15
Explanation:
Tilt of node 3 : |0-0| = 0 (no children)
Tilt of node 5 : |0-0| = 0 (no children)
Tilt of node 7 : |0-0| = 0 (no children)
Tilt of node 2 : |3-5| = 2 (left subtree is just left child, so sum is 3; right subtree is just right child, so sum is 5)
Tilt of node 9 : |0-7| = 7 (no left child, so sum is 0; right subtree is just right child, so sum is 7)
Tilt of node 4 : |(3+5+2)-(9+7)| = |10-16| = 6 (left subtree values are 3, 5, and 2, which sums to 10; right subtree values are 9 and 7, which sums to 16)
Sum of every tilt : 0 + 0 + 0 + 2 + 7 + 6 = 15
Example 3:
Input: root = [21,7,14,1,1,2,2,3,3]
Output: 9
Constraints:
The number of nodes in the tree is in the range [0, 104].
-1000 <= Node.val <= 1000
思路一：用一个全局变量来储存所有的节点的倾斜值，递归计算树的每个节点
private static int TILT_SUM = 0;
public static int findTilt(TreeNode root) {
TILT_SUM = 0;
findTiltRecursive(root);
return TILT_SUM;
}
public static int findTiltRecursive(TreeNode root) {
if (root == null) {
return 0;
}
int leftTilt = findTiltRecursive(root.left);
int rightTilt = findTiltRecursive(root.right);
int abs = Math.abs(leftTilt - rightTilt);
TILT_SUM += abs;
return leftTilt + rightTilt + root.val;
}