LC.P623[在二叉树中增加一行]

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 addOneRow(TreeNode root, int val, int depth) {
        if (depth == 1) return new TreeNode(val, root, null);
        int currentDepth = 1;
        Deque<TreeNode> queue = new ArrayDeque<>();
        if (root != null) queue.offer(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            while (size-- > 0) {
                TreeNode p = queue.poll();
                if (depth - 1 == currentDepth) {
                    p.left = new TreeNode(val, p.left, null);
                    p.right = new TreeNode(val, null, p.right);
                } else {
                    if (p.left != null) queue.offer(p.left);
                    if (p.right != null) queue.offer(p.right);
                }
            }
            ++currentDepth;
        }
        return root;
    }
}
  • 时间复杂度；$O(n)$
  • 空间复杂度：$O(n)$

方法二：DFS

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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 {

    int depth, val;

    public TreeNode addOneRow(TreeNode root, int val, int depth) {
        if (depth == 1) return new TreeNode(val, root, null);
        this.depth = depth;
        this.val = val;
        dfs(root, 1);
        return root;
    }
    
    private void dfs(TreeNode root, int cur) {
        if (root == null) return;
        if (cur == depth - 1) {
            TreeNode a = new TreeNode(val), b = new TreeNode(val);
            a.left = root.left;
            b.right = root.right;
            root.left = a;
            root.right = b;
        } else {
            dfs(root.left, cur + 1);
            dfs(root.right, cur + 1);
        }
    }
}
  • 时间复杂度；$O(n)$
  • 空间复杂度：$O(n)$