leetcode 623. Add One Row to Tree

Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with value v at the given depth d. The root node is at depth 1.

The adding rule is: given a positive integer depth d, for each NOT null tree nodes N in depth d-1, create two tree nodes with value v as N's left subtree root and right subtree root. And N's original left subtree should be the left subtree of the new left subtree root, its original right subtree should be the right subtree of the new right subtree root. If depth d is 1 that means there is no depth d-1 at all, then create a tree node with value v as the new root of the whole original tree, and the original tree is the new root's left subtree.

Example 1:

Input: A binary tree as following:       4     /   \    2     6   / \   /   3   1 5   v = 1d = 2Output:        4      / \     1   1    /     \   2       6  / \     /  3   1   5   

Example 2:

Input: A binary tree as following:      4     /       2       / \     3   1    v = 1d = 3Output:       4     /       2   / \      1   1 /     \  3       1

Note:

1. The given d is in range [1, maximum depth of the given tree + 1].
2. The given binary tree has at least one tree node.

这一题也是水题，DFS就OK，大神的DFS和我的思路是一样的。

package leetcode;public class Add_One_Row_to_Tree_623 {public TreeNode addOneRow(TreeNode root, int v, int d) {if(d==1){TreeNode newRoot=new TreeNode(v);newRoot.left=root;return newRoot;}else{int deep=d-1;addRow(root, v, deep, 1);}return root;}public void addRow(TreeNode node,int v,int deep,int thisDeep){if(node==null){return;}if(deep==thisDeep){TreeNode leftTemp=node.left;TreeNode rightTemp=node.right;node.left=new TreeNode(v);node.right=new TreeNode(v);node.left.left=leftTemp;node.right.right=rightTemp;return;}addRow(node.left, v, deep, thisDeep+1);addRow(node.right, v, deep, thisDeep+1);}public static void main(String[] args) {// TODO Auto-generated method stubAdd_One_Row_to_Tree_623 a=new Add_One_Row_to_Tree_623();TreeNode root=new TreeNode(4);root.left=new TreeNode(2);root.right=new TreeNode(6);root.left.left=new TreeNode(3);root.left.right=new TreeNode(1);root.right.left=new TreeNode(5);root=a.addOneRow(root, 1, 2);System.out.println("1");}}

也有大神用BFS，首先队列出队列进，使队列中最终只剩下d-1那一行的TreeNode，再针对这一行的node进行添加子元素的操作。

//BFS, find the d-1th row and add new children to each of them   public TreeNode addOneRow(TreeNode root, int v, int d) {        if (d == 1) {            TreeNode newroot = new TreeNode(v);            newroot.left = root;            return newroot;        }        LinkedList<TreeNode> queue = new LinkedList<>();        queue.add(root);        for (int i = 0; i < d-2; i++) {            int size = queue.size();            for (int j = 0; j < size; j++) {                TreeNode t = queue.poll();                if (t.left != null) queue.add(t.left);                if (t.right != null) queue.add(t.right);            }        }        while (!queue.isEmpty()) {            TreeNode t = queue.poll();            TreeNode tmp = t.left;            t.left = new TreeNode(v);            t.left.left = tmp;            tmp = t.right;            t.right = new TreeNode(v);            t.right.right = tmp;        }        return root;    }