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623. Add One Row to Tree

Description

Given the root of a binary tree and two integers val and depth, add a row of nodes with value val at the given depth depth.

Note that the root node is at depth 1.

The adding rule is:

  • Given the integer depth, for each not null tree node cur at the depth depth - 1, create two tree nodes with value val as cur's left subtree root and right subtree root.
  • cur's original left subtree should be the left subtree of the new left subtree root.
  • cur's original right subtree should be the right subtree of the new right subtree root.
  • If depth == 1 that means there is no depth depth - 1 at all, then create a tree node with value val as the new root of the whole original tree, and the original tree is the new root's left subtree.

 

Example 1:

Input: root = [4,2,6,3,1,5], val = 1, depth = 2
Output: [4,1,1,2,null,null,6,3,1,5]

Example 2:

Input: root = [4,2,null,3,1], val = 1, depth = 3
Output: [4,2,null,1,1,3,null,null,1]

 

Constraints:

  • The number of nodes in the tree is in the range [1, 104].
  • The depth of the tree is in the range [1, 104].
  • -100 <= Node.val <= 100
  • -105 <= val <= 105
  • 1 <= depth <= the depth of tree + 1

Solutions

  • /**
     * 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);
        }
    }
    
  • /**
     * 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);
        }
    };
    
  • # 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: Optional[TreeNode], val: int, depth: int
        ) -> Optional[TreeNode]:
            def dfs(root, d):
                if root is None:
                    return
                if d == depth - 1:
                    root.left = TreeNode(val, root.left, None)
                    root.right = TreeNode(val, None, root.right)
                    return
                dfs(root.left, d + 1)
                dfs(root.right, d + 1)
    
            if depth == 1:
                return TreeNode(val, root)
            dfs(root, 1)
            return root
    
    
  • /**
     * 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
    }
    
  • /**
     * Definition for a binary tree node.
     * class TreeNode {
     *     val: number
     *     left: TreeNode | null
     *     right: TreeNode | null
     *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
     *         this.val = (val===undefined ? 0 : val)
     *         this.left = (left===undefined ? null : left)
     *         this.right = (right===undefined ? null : right)
     *     }
     * }
     */
    
    function addOneRow(root: TreeNode | null, val: number, depth: number): TreeNode | null {
        if (depth === 1) {
            return new TreeNode(val, root);
        }
    
        const queue = [root];
        for (let i = 1; i < depth - 1; i++) {
            const n = queue.length;
            for (let j = 0; j < n; j++) {
                const { left, right } = queue.shift();
                left && queue.push(left);
                right && queue.push(right);
            }
        }
        for (const node of queue) {
            node.left = new TreeNode(val, node.left);
            node.right = new TreeNode(val, null, node.right);
        }
        return root;
    }
    
    

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