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 and sum = 22
,
5 / 4 8 / / 11 13 4 / 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2
which sum is 22.
dfs method:
public class Solution { public boolean hasPathSum(TreeNode root, int sum) { if(root == null) return false; return dfs(root, 0, sum); } private boolean dfs (TreeNode node, int accu, int sum) { accu += node.val; if(node.left == null && node.right == null && accu == sum) return true; else { if(node.left != null) { if(dfs(node.left, accu, sum)) return true; } if(node.right != null) { if(dfs(node.right, accu, sum)) return true; } } accu -= node.val; return false; } }
another code:
/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution { /** * This problem can be solved either by dfs or bfs.<br> * @param root * @param sum * @return */ public boolean hasPathSum(TreeNode root, int sum) { boolean exists = false; if(root != null){ Map<TreeNode, Integer> sumAtNode = new HashMap<TreeNode, Integer>(); sumAtNode.put(root, root.val); Deque<TreeNode> nodes = new LinkedList<TreeNode>(); nodes.add(root); while(nodes.peek()!= null){ TreeNode aNode = nodes.poll(); if(aNode.left == null && aNode.right == null){ if(sumAtNode.get(aNode) == sum){ exists = true; break; } } else{ int pareValue = sumAtNode.get(aNode); if(aNode.left != null){ sumAtNode.put(aNode.left, aNode.left.val + pareValue); nodes.add(aNode.left); } if(aNode.right != null){ sumAtNode.put(aNode.right, aNode.right.val + pareValue); nodes.add(aNode.right); } } } } return exists; } }