Title Description
- An input and a binary value, request from the root to the leaf node, and the value equal to the path
- Deformation depth-first search
Problem-solving ideas
Deformation depth-first search
Two solutions python
Recursive
# -*- coding:utf-8 -*-
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# 返回二维列表，内部每个列表表示找到的路径
def FindPath(self, root, expectNumber):
# write code here
ret = []
if not root:
return ret
path = [root]
sums = [root.val]
def dfs(root):
if root.left:
path.append(root.left)
sums.append(sums[-1]+root.left.val)
dfs(root.left)
if root.right:
path.append(root.right)
sums.append(sums[-1] + root.right.val)
dfs(root.right)
if not root.left and not root.right:
if sums[-1] == expectNumber:
ret.append([p.val for p in path])
path.pop()
sums.pop()
dfs(root)
return ret
if __name__ == '__main__':
t = Tree()
t.construct_tree([1, 3, 6, 4, 3, 1, 1])
print(find_path(t.root, 8)) # [[1, 3, 4], [1, 6, 1], [1, 6, 1]]
Backtracking
class Solution:
def FindPath(self, root, expectNumber):
def subFindPath(root):
if root:
b.append(root.val)
if not root.right and not root.left and sum(b) == expectNumber:
a.append(b[:])
else:
subFindPath(root.left),subFindPath(root.right)
b.pop()
a, b = [], []
subFindPath(root)
return a
