Binary Tree Backtracking¶
Table of Contents¶
- 257. Binary Tree Paths (Easy)
- 113. Path Sum II (Medium)
- 437. Path Sum III (Medium)
257. Binary Tree Paths¶
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LeetCode | LeetCode CH (Easy)
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Tags: string, backtracking, tree, depth first search, binary tree
257. Binary Tree Paths - Python Solution
from typing import List, Optional
from binarytree import Node as TreeNode
from binarytree import build
# Recursive
def binaryTreePaths(root: Optional[TreeNode]) -> List[str]:
res = []
def dfs(node, path):
if not node:
return
path += str(node.val)
if not node.left and not node.right:
res.append(path)
return
path += "->"
dfs(node.left, path)
dfs(node.right, path)
dfs(root, "")
return res
root = build([1, 2, 3, None, 5])
print(root)
# __1
# / \
# 2 3
# \
# 5
print(binaryTreePaths(root)) # ['1->2->5', '1->3']
113. Path Sum II¶
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LeetCode | LeetCode CH (Medium)
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Tags: backtracking, tree, depth first search, binary tree
437. Path Sum III¶
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LeetCode | LeetCode CH (Medium)
-
Tags: tree, depth first search, binary tree
437. Path Sum III - C++ Solution#include <iostream> #include <unordered_map> #include <vector> using namespace std; struct TreeNode { int val; TreeNode *left; TreeNode *right; TreeNode() : val(0), left(nullptr), right(nullptr) {} TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} }; class Solution { public: int pathSum(TreeNode *root, int targetSum) { int res = 0; unordered_map<long long, int> cnt{{0, 1}}; auto dfs = [&](auto &&self, TreeNode *node, long long cur) { if (!node) return; cur += node->val; if (cnt.find(cur - targetSum) != cnt.end()) res += cnt[cur - targetSum]; cnt[cur]++; self(self, node->left, cur); self(self, node->right, cur); cnt[cur]--; }; dfs(dfs, root, 0); return res; } }; int main() { Solution s; { TreeNode *root = new TreeNode(10); root->left = new TreeNode(5); root->right = new TreeNode(-3); root->left->left = new TreeNode(3); root->left->right = new TreeNode(2); root->right->right = new TreeNode(11); root->left->left->left = new TreeNode(3); root->left->left->right = new TreeNode(-2); root->left->right->right = new TreeNode(1); cout << s.pathSum(root, 8) << endl; // 3 } { TreeNode *root = new TreeNode(5); root->left = new TreeNode(4); root->right = new TreeNode(8); root->left->left = new TreeNode(11); root->right->left = new TreeNode(13); root->right->right = new TreeNode(4); root->left->left->left = new TreeNode(7); root->left->left->right = new TreeNode(2); root->right->right->left = new TreeNode(5); root->right->right->right = new TreeNode(1); cout << s.pathSum(root, 22) << endl; // 3 } return 0; }