437. Path Sum III.md

437. Path Sum III

Question:

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:

• Method 1: dfs + bfs

  /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode(int x) { val = x; }
   * }
   */
  class Solution {
      private int result;
      public int pathSum(TreeNode root, int sum) {
          if(root == null) return 0;
          Queue<TreeNode> q = new LinkedList<>();
          q.offer(root);
          while(!q.isEmpty()){
              TreeNode node = q.poll();
              dfs(node, 0, sum);
              if(node.left != null) q.offer(node.left);
              if(node.right != null) q.offer(node.right);
          }
          return result;
      }
      private void dfs(TreeNode node, int cur, int sum){
          if(node == null) return;
          if(cur + node.val == sum) result++;
          dfs(node.left, cur + node.val, sum);
          dfs(node.right, cur + node.val, sum);
      }
  }