LeetCode-in-Java
2641. Cousins in Binary Tree II
Medium
Given the root of a binary tree, replace the value of each node in the tree with the sum of all its cousins’ values.
Two nodes of a binary tree are cousins if they have the same depth with different parents.
Return the root of the modified tree.
Note that the depth of a node is the number of edges in the path from the root node to it.
Example 1:
Input: root = [5,4,9,1,10,null,7]
Output: [0,0,0,7,7,null,11]
Explanation: The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
- Node with value 5 does not have any cousins so its sum is 0.
- Node with value 4 does not have any cousins so its sum is 0.
- Node with value 9 does not have any cousins so its sum is 0.
- Node with value 1 has a cousin with value 7 so its sum is 7.
- Node with value 10 has a cousin with value 7 so its sum is 7.
- Node with value 7 has cousins with values 1 and 10 so its sum is 11.
Example 2:
Input: root = [3,1,2]
Output: [0,0,0]
Explanation: The diagram above shows the initial binary tree and the binary tree after changing the value of each node.
- Node with value 3 does not have any cousins so its sum is 0.
- Node with value 1 does not have any cousins so its sum is 0.
- Node with value 2 does not have any cousins so its sum is 0.
Constraints:
- The number of nodes in the tree is in the range
[1, 105]. 1 <= Node.val <= 104
Solution
import com_github_leetcode.TreeNode;
import java.util.ArrayList;
import java.util.List;
/*
* 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;
* }
* }
*/
public class Solution {
private List<Integer> horizontalSum;
private void traverse(TreeNode root, int depth) {
if (root == null) {
return;
}
if (depth < horizontalSum.size()) {
horizontalSum.set(depth, horizontalSum.get(depth) + root.val);
} else {
horizontalSum.add(root.val);
}
traverse(root.left, depth + 1);
traverse(root.right, depth + 1);
}
private void traverse1(TreeNode root, int depth) {
if (root == null) {
return;
}
if (depth > 0) {
int sum = 0;
if (root.left != null) {
sum += root.left.val;
}
if (root.right != null) {
sum += root.right.val;
}
if (root.left != null) {
root.left.val = horizontalSum.get(depth + 1) - sum;
}
if (root.right != null) {
root.right.val = horizontalSum.get(depth + 1) - sum;
}
}
traverse1(root.left, depth + 1);
traverse1(root.right, depth + 1);
}
public TreeNode replaceValueInTree(TreeNode root) {
horizontalSum = new ArrayList<>();
root.val = 0;
if (root.left != null) {
root.left.val = 0;
}
if (root.right != null) {
root.right.val = 0;
}
traverse(root, 0);
traverse1(root, 0);
return root;
}
}