112 Path Sum
112. Path Sum
1. Question
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.
For example:
Given the below binary tree andsum = 22
5
/ \
4 8
/ / \
11 13 4
/ \ \
7 2 1
return true, as there exist a root-to-leaf path5->4->11->2
which sum is 22.
2. Implementation
(1) Recursion
public class Solution {
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) return false;
if(root.left == null && root.right == null) return sum == root.val;
return hasPathSum(root.left, sum - root.val) || hasPathSum(root.right, sum - root.val);
}
}
(2) Iteration
public boolean hasPathSum(TreeNode root, int sum) {
if (root == null) {
return false;
}
Stack<TreeNode> nodes = new Stack<>();
Stack<Integer> pathSums = new Stack<>();
nodes.push(root);
pathSums.push(root.val);
while (!nodes.isEmpty()) {
TreeNode curNode = nodes.pop();
int curSum = pathSums.pop();
if (curNode.left == null && curNode.right == null && curSum == sum) {
return true;
}
if (curNode.left != null) {
nodes.push(curNode.left);
pathSums.push(curSum + curNode.left.val);
}
if (curNode.right != null) {
nodes.push(curNode.right);
pathSums.push(curSum + curNode.right.val);
}
}
return false;
}
3. Time & Space Complexity
Recursion: 时间复杂度: O(n), 空间复杂度:O(n)
Iteration: 时间复杂度: O(n), 空间复杂度: O(n)
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