LC.P2641[二叉树的堂兄弟节点II]

方法一：BFS

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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 {
    public TreeNode replaceValueInTree(TreeNode root) {
        root.val = 0;
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            // 统计每一层的节点和
            int sum = 0;
            for (TreeNode node : queue) {
                if (node.left != null) {
                    sum += node.left.val;
                }
                if (node.right != null) {
                    sum += node.right.val;
                }
            }
            int size = queue.size();
            while (size-- > 0) {
                TreeNode node = queue.poll();
                int sub = (node.left == null ? 0 : node.left.val) +
                        (node.right == null ? 0 : node.right.val);
                if (node.left != null) {
                    node.left.val = sum - sub;
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    node.right.val = sum - sub;
                    queue.offer(node.right);
                }
            }
        }
        return root;
    }
}
  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(n)$