I

描述

给定一个二叉树，找出所有路径中各节点相加总和等于给定 目标值 的路径。

一个有效的路径，指的是从根节点到叶节点的路径。

样例

给定一个二叉树，和 目标值 = 5:

 1
/ \

2 4
/
2 3
返回：

[
[1, 2, 2],
[1, 4]
]

代码

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""

class Solution:
    """
    @param root: The root of binary tree.
    @return: An integer
    """
    def maxPathSum(self, root):
        # write your code here
        self.maxPath = root.val
        self.Dfs(root)
        return self.maxPath
    
    def Dfs(self, root):
        # TODO: write code...
        if root is None:
            return 0
        left = max(0, self.Dfs(root.left))
        right = max(0, self.Dfs(root.right))
        leap = left + root.val + right
        if leap > self.maxPath:
            self.maxPath = leap
        return max(left + root.val, right + root.val)

题目来源

https://www.lintcode.com/problem/binary-tree-path-sum-i/description

II

描述

给出一棵二叉树，寻找一条路径使其路径和最大，路径可以在任一节点中开始和结束（路径和为两个节点之间所在路径上的节点权值之和）

样例

给出一棵二叉树：

   1
  / \
 2   3

返回 6

代码

这个写法性能差一点。

"""
Definition of TreeNode:
class TreeNode:
    def __init__(self, val):
        self.val = val
        self.left, self.right = None, None
"""


class Solution:
    """
    @param: root: the root of binary tree
    @param: target: An integer
    @return: all valid paths
    """
    def binaryTreePathSum2(self, root, target):
        # write your code here
        self.result = []
        if root is None:
            return self.result
        self.target = target
        local = []
        self.Dfs(root, 0, local)
        return self.result
        
    def Dfs(self, root, sumPath, local):
        if root is None:
            return
        local.append(root.val)
        sumPath += root.val
        if sumPath == self.target:
            self.result.append(list(local))
        new_local = local[:]
        sumPath_local = sumPath
        while(len(new_local)>1):
            sumPath_local -= new_local[0]
            del new_local[0]
            if sumPath_local == self.target:
                self.result.append(list(new_local))
        self.Dfs(root.left, sumPath, local)
        self.Dfs(root.right, sumPath, local)
        local.pop()

这个代码的性能要好一些，但是算法都一样。

class Solution:
    """
    @param: root: the root of binary tree
    @param: target: An integer
    @return: all valid paths
    """
    def binaryTreePathSum2(self, root, target):
        # write your code here
        self.result = []
        if root is None:
            return self.result
        self.target = target
        local = []
        self.Dfs(root, 0, local)
        return self.result
        
    def Dfs(self, root, sumPath, local):
        if root is None:
            return
        local.append(root.val)
        sumPath += root.val
        if sumPath == self.target:
            self.result.append(list(local))
        for i in range(1, len(local)):
            new_local = local[i:len(local)]
            print(new_local)
            if sum(new_local) == self.target:
                self.result.append(list(new_local))
        self.Dfs(root.left, sumPath, local)
        self.Dfs(root.right, sumPath, local)
        local.pop()

到达每个子节点都有一个“回头观测”的过程，

 for i in range(1, len(local)):
            new_local = local[i:len(local)]
            print(new_local)
            if sum(new_local) == self.target:
                self.result.append(list(new_local))

把先进入的节点依次撇除然后再计算是否符合结果。

题目来源

https://www.lintcode.com/problem/binary-tree-path-sum-ii/description