LeetCode.623 Add One Row to Tree

题目：

Given the root of a binary tree, then value v and depth d, you need to add a row of nodes with valuev 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 nodesN 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, itsoriginal right subtree should be the right subtree of the new right subtree root. If depthd is 1 that means there is no depth d-1 at all, then create a tree node with valuev 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:

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

分析（原创）：

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {public TreeNode addOneRow(TreeNode root, int v, int d) {//给定二叉树，在指定的深度d添加指定的值v,并且v作为新的左右子树节点//思路：记录当前节点的深度，当count==d-1时，则进行节点生成，以及对新节点继承父节点的值。//注意：如果d为1，则只需要把根节点作为其右子树返回即可return addOneRow(root,v,d,1);}public TreeNode addOneRow(TreeNode root, int v, int d,int lastDepth){if(root==null) return root;if(d==1){TreeNode newRoot=new TreeNode(v);newRoot.left=root;return newRoot;}if(curDepth(root,lastDepth)==d-1){//如果刚好判断到d-1层，则后面节点进行拼接TreeNode tempLeft=new TreeNode(v);if(root.left!=null){tempLeft.left=root.left;}TreeNode tempRight=new TreeNode(v);if(root.right!=null){tempRight.right=root.right;}root.left=tempLeft;root.right=tempRight;}else{root.left=addOneRow(root.left,v,d,lastDepth+1);root.right=addOneRow(root.right,v,d,lastDepth+1);}return root;}//找到当前节点的层数public int curDepth(TreeNode root,int lastDepth){return root==null?1+lastDepth:lastDepth;}
}

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