LeetCode:Path Sum

Path Sum


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.

BFS:

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root==null) return false;
        
        LinkedList<TreeNode> nodes = new LinkedList<TreeNode>();
        LinkedList<Integer> accSums = new LinkedList<Integer>();
        nodes.add(root);
        accSums.add(root.val);
        
        while(!nodes.isEmpty()){
            TreeNode node = nodes.poll();
            Integer accSum = accSums.poll();
            if(node.left==null && node.right==null && accSum==sum)
                return true;
            if(node.left!=null){
                nodes.add(node.left);
                accSums.add(accSum+node.left.val);
            }
            if(node.right!=null){
                nodes.add(node.right);
                accSums.add(accSum+node.right.val);
            }
        }
        return false;
    }
}

DFS:

public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(root==null) return false;
        
        Stack<TreeNode> nodes = new Stack<TreeNode>();
        Stack<Integer> accSums = new Stack<Integer>();
        nodes.add(root);
        accSums.add(root.val);
        
        while(!nodes.isEmpty()){            
            TreeNode node = nodes.pop();
            Integer accSum = accSums.pop();
            if(node.left==null && node.right==null && accSum==sum)
                return true;
            
            if(node.right!=null){
                nodes.add(node.right);
                accSums.add(accSum+node.right.val);
            }
            
            if(node.left!=null){
                nodes.add(node.left);
                accSums.add(accSum+node.left.val);
            }
            
           
        }
        return false;
    }
}