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623. 在二叉树中增加一行

English Version

题目描述

给定一个二叉树的根 root 和两个整数 val 和 depth ，在给定的深度 depth 处添加一个值为 val 的节点行。

注意，根节点 root 位于深度 1 。

加法规则如下:

• 给定整数 depth，对于深度为 depth - 1 的每个非空树节点 cur ，创建两个值为 val 的树节点作为 cur 的左子树根和右子树根。
• cur 原来的左子树应该是新的左子树根的左子树。
• cur 原来的右子树应该是新的右子树根的右子树。
• 如果 depth == 1 意味着 depth - 1 根本没有深度，那么创建一个树节点，值 val 作为整个原始树的新根，而原始树就是新根的左子树。

 

示例 1:

输入: root = [4,2,6,3,1,5], val = 1, depth = 2
输出: [4,1,1,2,null,null,6,3,1,5]

示例 2:

输入: root = [4,2,null,3,1], val = 1, depth = 3
输出:  [4,2,null,1,1,3,null,null,1]

 

提示:

• 节点数在 [1, 104] 范围内
• 树的深度在 [1, 104]范围内
• -100 <= Node.val <= 100
• -105 <= val <= 105
• 1 <= depth <= the depth of tree + 1

解法

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def addOneRow(self, root: TreeNode, val: int, depth: int) -> TreeNode:
        def dfs(root, d):
            if root is None:
                return
            if d == depth - 1:
                l = TreeNode(val=val, left=root.left)
                r = TreeNode(val=val, right=root.right)
                root.left, root.right = l, r
                return
            dfs(root.left, d + 1)
            dfs(root.right, d + 1)

        if depth == 1:
            return TreeNode(val=val, left=root)
        dfs(root, 1)
        return root

Java

/**
 * 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 {
    private int val;
    private int depth;

    public TreeNode addOneRow(TreeNode root, int val, int depth) {
        if (depth == 1) {
            return new TreeNode(val, root, null);
        }
        this.val = val;
        this.depth = depth;
        dfs(root, 1);
        return root;
    }

    private void dfs(TreeNode root, int d) {
        if (root == null) {
            return;
        }
        if (d == depth - 1) {
            TreeNode l = new TreeNode(val, root.left, null);
            TreeNode r = new TreeNode(val, null, root.right);
            root.left = l;
            root.right = r;
            return;
        }
        dfs(root.left, d + 1);
        dfs(root.right, d + 1);
    }
}

C++

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int val;
    int depth;

    TreeNode* addOneRow(TreeNode* root, int val, int depth) {
        if (depth == 1) return new TreeNode(val, root, nullptr);
        this->val = val;
        this->depth = depth;
        dfs(root, 1);
        return root;
    }

    void dfs(TreeNode* root, int d) {
        if (!root) return;
        if (d == depth - 1)
        {
            auto l = new TreeNode(val, root->left, nullptr);
            auto r = new TreeNode(val, nullptr, root->right);
            root->left = l;
            root->right = r;
            return;
        }
        dfs(root->left, d + 1);
        dfs(root->right, d + 1);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func addOneRow(root *TreeNode, val int, depth int) *TreeNode {
	if depth == 1 {
		return &TreeNode{Val: val, Left: root}
	}
	var dfs func(root *TreeNode, d int)
	dfs = func(root *TreeNode, d int) {
		if root == nil {
			return
		}
		if d == depth-1 {
			l, r := &TreeNode{Val: val, Left: root.Left}, &TreeNode{Val: val, Right: root.Right}
			root.Left, root.Right = l, r
			return
		}
		dfs(root.Left, d+1)
		dfs(root.Right, d+1)
	}
	dfs(root, 1)
	return root
}

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