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Copy path112_Path_Sum.java
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126 lines (115 loc) · 3.33 KB
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/*
* 112. Path Sum
* Target: Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that
* adding up all the values along the path equals the given sum.
* Note: A leaf is a node with no children.
* Difficulty：Easy
* Classification：Tree, DFS
*/
/*
* Solution 1
* 2019-10-29 Runtime: 4 ms
* Algorithm: => DFS
* Time Complexity: ?, Space Conplexity: O(1)
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
Stack<TreeNode> stack = new Stack<>();
Stack<Integer> stackSum = new Stack<>();
TreeNode cur = root;
stack.push(root);
stackSum.push(root.val);
int pathSum = 0;
while (cur != null || !stack.isEmpty()) {
while (cur != null) {
stack.push(cur);
pathSum += cur.val;
stackSum.push(pathSum);
cur = cur.left;
}
cur = stack.pop();
pathSum = stackSum.pop();
if (cur.left == null && cur.right == null && pathSum == sum) {
return true;
}
cur = cur.right;
}
return false;
}
}
/*
* Solution 2
* 2019-10-29 Runtime: 0 ms
* Algorithm: => Recursion
* Time Complexity: ?, Space Conplexity: O(1)
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
return pathSumHelper(root, sum);
}
private boolean pathSumHelper(TreeNode cur, int sum) {
if (cur.left == null && cur.right == null) {
return cur.val == sum;
}
if (cur.left != null && cur.right == null) {
return pathSumHelper(cur.left, sum - cur.val);
}
if (cur.right != null && cur.left == null) {
return pathSumHelper(cur.right, sum - cur.val);
}
return pathSumHelper(cur.left, sum - cur.val) || pathSumHelper(cur.right, sum - cur.val);
}
}
/*
* Solution 3
* 2019-10-29 Runtime: 1 ms
* Algorithm: => BFS
* Time Complexity: ?, Space Conplexity: O(1)
*/
class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
Queue<TreeNode> q = new LinkedList<>();
Queue<Integer> qSum = new LinkedList<>();
q.offer(root);
qSum.offer(root.val);
TreeNode cur = root;
int pathSum = root.val;
while (!q.isEmpty()) {
int level = q.size();
for (int i = 0; i < level; i++) {
cur = q.poll();
pathSum = qSum.poll();
if (cur.left == null && cur.right == null && pathSum == sum) {
return true;
}
if (cur.left != null) {
q.offer(cur.left);
qSum.offer(pathSum + cur.left.val);
}
if (cur.right != null) {
q.offer(cur.right);
qSum.offer(pathSum + cur.right.val);
}
}
}
return false;
}
}