leetcode 814. Binary Tree Pruning

leetcode 814. Binary Tree Pruning

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原题地址：https://leetcode.com/problems/binary-tree-pruning/

题目

We are given the head node root of a binary tree, where additionally every node’s value is either a 0 or a 1.

Return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

(Recall that the subtree of a node X is X, plus every node that is a descendant of X.)

Example 1:
Input: [1,null,0,0,1]
Output: [1,null,0,null,1]

Explanation: 
Only the red nodes satisfy the property "every subtree not containing a 1".
The diagram on the right represents the answer.

Example 2:
Input: [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]

Example 3:
Input: [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]

Note:

• The binary tree will have at most 100 nodes.
• The value of each node will only be 0 or 1.

python代码1

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

class Solution:
    root1 = TreeNode(0)
    def pruneTree(self, root):
        """
        :type root: TreeNode
        :rtype: TreeNode
        """
        if root is not None:
            self.root1 = root

        def ok(node, node1):
            if node == None:
                node1 = None
            else:
                node1.val = node.val
                if self.hasOne(node.left):
                    node1.left = node.left
                    ok(node.left, node1.left)
                    ok(node.right, node1.right)
                else:
                    node1.left = None

                if self.hasOne(node.right):
                    node1.right = node.right
                    ok(node.left, node1.left)
                    ok(node.right, node1.right)
                else:
                    node1.right = None                                    
        ok(root, self.root1)

        return self.root1    

    def hasOne(self, node):
        if node is None:
            return False
        if node.val == 1:
            return True
        elif node.val == 0:
            return self.hasOne(node.left) or self.hasOne(node.right)

python代码2

def pruneTree(self, root):
    if not root: return None
    root.left = self.pruneTree(root.left)
    root.right = self.pruneTree(root.right)
    if not root.left and not root.right and not root.val: return None
    return root

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