112. Path Sum

Problem

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.

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.

Solution Recursive

class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if (root == null) return false;        //has to be a root-to-leaf path        if (root.val == sum && root.left == null && root.right == null) return true;        return hasPathSum(root.left, sum-root.val) || hasPathSum(root.right, sum-root.val);    }}

Update 2018-11

Solution Iterative

class Solution {    public boolean hasPathSum(TreeNode root, int sum) {        if (root == null) return false;        Deque
    
      nodestack = new ArrayDeque<>();        Deque
     
       sumstack = new ArrayDeque<>();        nodestack.push(root);        sumstack.push(sum);        while (!nodestack.isEmpty()) {            TreeNode curNode = nodestack.pop();            int curSum = sumstack.pop();            if (curNode.left == null && curNode.right == null && curNode.val == curSum) return true;            if (curNode.right != null) {                nodestack.push(curNode.right);                sumstack.push(curSum-curNode.val);            }            if (curNode.left != null) {                nodestack.push(curNode.left);                sumstack.push(curSum-curNode.val);            }        }        return false;    }}
     
    

Path Sum II

Problem

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.

Note: A leaf is a node with no children.

Example:

Given the below binary tree and sum = 22,

5     / \    4   8   /   / \  11  13  4 /  \    / \7    2  5   1

Return:

[   [5,4,11,2],   [5,8,4,5]]

Solution

class Solution {    public List
    
     
      > pathSum(TreeNode root, int sum) {        List
      
       
        > res = new ArrayList<>();        if (root == null) return res;        dfs(root, sum, new ArrayList
        
         (), res);        return res;    }    private void dfs(TreeNode root, int sum, List
         
           temp, List
          
           
            > res) { if (sum == root.val && root.left == null && root.right == null) { temp.add(root.val); List
            
              save = new ArrayList<>(temp); //important res.add(save); temp.remove(temp.size()-1); } else { temp.add(root.val); if (root.left != null) dfs(root.left, sum-root.val, temp, res); if (root.right != null) dfs(root.right, sum-root.val, temp, res); temp.remove(temp.size()-1); } }}
            
           
          
         
        
       
      
     
    

Path Sum III

Problem

You are given a binary tree in which each node contains an integer value.

Find the number of paths that sum to a given value.

The path does not need to start or end at the root or a leaf, but it must go downwards (traveling only from parent nodes to child nodes).

The tree has no more than 1,000 nodes and the values are in the range -1,000,000 to 1,000,000.

Example:

root = [10,5,-3,3,2,null,11,3,-2,null,1], sum = 8

10     /  \    5   -3   / \    \  3   2   11 / \   \3  -2   1

Return 3. The paths that sum to 8 are:

1. 5 -> 3
2. 5 -> 2 -> 1
3. -3 -> 11

Solution

class Solution {    public int pathSum(TreeNode root, int sum) {        if (root == null) return 0;        return helper(root, sum) + pathSum(root.left, sum) + pathSum(root.right, sum);    }    private int helper(TreeNode root, int sum) {                int count = 0;        if (sum == root.val) count++;        if (root.left != null) count += helper(root.left, sum-root.val);        if (root.right != null) count += helper(root.right, sum-root.val);        return count;    }}