Path Sum II
Path Sum II
题目
Given a binary tree and a sum, find all root-to-leaf paths where each path’s sum equals the given sum.
For example:
Given the below binary tree and sum = 22,
5
/ \
4 8
/ / \
11 13 4
/ \ / \
7 2 5 1
return
[
[5,4,11,2],
[5,8,4,5]
]
思路
跟path sum思路一样，只不过这里要保存经过的路径。
解题
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32/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> result;
vector<int> cur;
pathSum(root,sum,cur,result);
return result;
}
void pathSum(TreeNode*root,int sum,vector<int>&cur,vector<vector<int>>& result){
if(!root)
return;
cur.push_back(root->val);
if(!root->left && !root->right){
if(root->val==sum){
result.push_back(cur);
}
}
pathSum(root->left,sum-root->val,cur,result);
pathSum(root->right,sum-root->val,cur,result);
cur.pop_back();
}
};
python版1
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27# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
# @param {TreeNode} root
# @param {integer} sum
# @return {integer[][]}
def pathSum(self, root, sum):
result=[]
cur=[]
self.hasPath(root,sum,cur,result)
return result
def hasPath(self,root,p,cur,result):
if not root:
return
cur.append(root.val)
if not root.left and not root.right:
if root.val==p:
result.append(cur)
self.hasPath(root.left,p-root.val,cur,result)
self.hasPath(root.right,p-root.val,cur,result)
cur.pop()
