力扣：623. 在二叉树中增加一行

最新推荐文章于 2025-12-02 18:04:47 发布

原创最新推荐文章于 2025-12-02 18:04:47 发布 · 139 阅读

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#leetcode #深度优先 #算法

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本文探讨了如何使用深度优先搜索（DFS）和广度优先搜索（BFS）算法在二叉树中实现添加新的一行的功能，通过实例展示了在不同深度条件下操作节点的技巧。

dfs

/**
 * 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 (root == null) return null;
        if (depth == 1) return new TreeNode(val, root, null);
        if (depth == 2) {
            root.left = new TreeNode(val, root.left, null); 
            root.right = new TreeNode(val, null, root.right);
        } else{
            root.left = addOneRow(root.left, val, depth - 1);
            root.right = addOneRow(root.right, val, depth - 1);
        }
        return root;
    }
}

bfs

/**
 * 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 (root == null){
            return null;
        }

        if (depth == 1){
            return new TreeNode(val, root, null);
        }
        List<TreeNode> list = new ArrayList<>();
        list.add(root);

        for (int i = 1; i < depth - 1; i++){
            List<TreeNode> temp = new ArrayList<>();
            for (TreeNode node : list) {
                if (node.left != null){
                    temp.add(node.left);
                }

                if (node.right != null){
                    temp.add(node.right);
                }
            }
            list = temp;
        }

        for (TreeNode node : list) {
            node.left = new TreeNode(val, node.left, null);
            node.right = new TreeNode(val, null,  node.right);
        }
        return root;
    }
}