Wins the Offer (XXIV): a binary tree and a path value

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First, the primer

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Second, the title

An input binary tree root node and an integer, the binary print values of nodes in the input path and all integers. Forming a path to the path definition begins from the root node of the tree down to the leaf node has been traversed nodes. (Note: the return value in the list, a large array Array front)

1, ideas

Two global variables result and tmp, result to hold the final result, tmp for storing temporary results.

Every traversal, we first root value stitching to tmp, and then determine whether the current root meet:

Left subtree is not empty
Right subtree is not empty
And whether the result of splicing and values are equal expectNumber

If the condition is satisfied, it will result in splicing tmp, otherwise, turn left and right sub-tree traversal. Note that, when the left and right sub-tree traversal, global variable tmp is backtracking to do, because if the path of a leaf node and does not meet the requirements we going back? This is not necessary to wipe off the leaf node and then back to the parent node to another child node access.

2, programming

python

Code implementation:

# -*- 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
        def subFindPath(root):
            if root:
                tmp.append(root.val)
                if not root.right and not root.left and sum(tmp) == expectNumber:
                    result.append(tmp[:])
                else:
                    subFindPath(root.left),subFindPath(root.right)
                # 回溯
                tmp.pop()
        result, tmp = [], []
        subFindPath(root)
        sorted(result, key=len, reverse=True)
        return result

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