112. Path Sum.md

112. Path Sum

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.

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.

Thinking:

• Method 1:递归
  • 这道题有一种特殊情况，当输入为[1, 2],sum为1.

    •   	1
      

    •   /
      

    • 2
  • 此时我们return false。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        return pathCount(root, 0, sum);
    }
    private static boolean pathCount(TreeNode node, int count, int sum){
        if(node == null){
            if(count == sum)
                return true;
            return false;
        }else{
            if(node.left == null)
                return pathCount(node.right, count + node.val, sum);
            else if(node.right == null)
                return pathCount(node.left, count + node.val, sum);
            else
                return pathCount(node.left, count + node.val, sum) || pathCount(node.right, count + node.val, sum);
        }
    }
}

二刷

简化了一刷时候的一些多余的逻辑判断。但是一刷时候的一些corner cases还是没有考虑进去。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if(root == null) return false;
        return hasPathSum(root, sum, 0);
    }
    private boolean hasPathSum(TreeNode node, int sum, int current){
        if(node.left == null && node.right == null){
            return (current + node.val) == sum;
        }else if(node.left != null && node.right == null){
            return hasPathSum(node.left, sum, current + node.val);
        }else if(node.left == null && node.right != null){
            return hasPathSum(node.right, sum, current + node.val);
        }else{
            return hasPathSum(node.left, sum, current + node.val) || hasPathSum(node.right, sum, current + node.val);
        }
    }
}

Third Time

• Method 1: DFS

   /**
    * Definition for a binary tree node.
    * public class TreeNode {
    *     int val;
    *     TreeNode left;
    *     TreeNode right;
    *     TreeNode(int x) { val = x; }
    * }
    */
   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 cur, int sum){
   		if(node.left == null && node.right == null){
   			return cur + node.val == sum;
   		}else{
   			if(node.left != null && dfs(node.left, cur + node.val, sum))   return true;
   			if(node.right != null && dfs(node.right, cur + node.val, sum))  return true;
   			return false;
   		}
   	}
   }