[LeetCode]Add One Row to Tree

内容简介：[LeetCode]Add One Row to Tree

题目描述：

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 = 1

d = 2

Output: 
       4
      / \
     1   1
    /     \
   2       6
  / \     / 
 3   1   5

Example 2:

Input: 
A binary tree as following:
      4
     /   
    2    
   / \   
  3   1    

v = 1

d = 3

Output: 
      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.

题目大意：

给定二叉树，根节点为第1层，在其第d层追加一行值为v的节点。将第d层各左孩子节点链接为新节点的左孩子，各右孩子节点链接为新节点的右孩子。

若d为1，则将原来的根节点链接为新节点的左孩子。

解题思路：

二叉树层次遍历

Python代码：

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def addOneRow(self, root, v, d):
        """
        :type root: TreeNode
        :type v: int
        :type d: int
        :rtype: TreeNode
        """
        if d == 1:
            nroot = TreeNode(v)
            nroot.left = root
            return nroot
        queue = [root]
        for x in range(1, d - 1):
            nqueue = []
            for node in queue:
                if node.left: nqueue.append(node.left)
                if node.right: nqueue.append(node.right)
            queue = nqueue
        for node in queue:
            left, right = node.left, node.right
            node.left, node.right = TreeNode(v), TreeNode(v)
            node.left.left, node.right.right = left, right
        return root

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