Path Sum II:
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] ]
Analysis:
Classic DFS search similar to the Question Path Sum I, in this problem we just add a vector<int> to store the result.
Details see code below:
Code(C++):
/**
* 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:
void ps(TreeNode* root, int sum, vector<int> path, vector<vector<int>> &res){
if (!root){
return;
}else{
path.push_back(root->val);
if (!root->left && !root->right && sum==root->val){
res.push_back(path);
path.clear();
}
ps(root->left, sum - root->val, path, res);
ps(root->right, sum - root->val, path, res);
}
}
vector<vector<int>> pathSum(TreeNode* root, int sum) {
vector<vector<int>> res;
vector<int>path;
ps(root, sum, path, res);
return res;
}
};
Code(Python):
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution(object):
def ps(self, root, sum, path, res):
if not root:
return
else:
path.append(root.val)
if not root.left and not root.right and sum == root.val:
res.append(path)
path = []
else:
self.ps(root.left, sum - root.val, path, res)
self.ps(root.right, sum - root.val, path, res)
def pathSum(self, root, sum):
"""
:type root: TreeNode
:type sum: int
:rtype: List[List[int]]
"""
res = []
path = []
self.ps(root, sum, path, res)
return res
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