【LeetCode】Path Sum

Path Sum

Total Accepted: 5082 Total Submissions: 17224 My Submissions

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.

用的bfs，AC以后一想，其实dfs也可以做。

dfs和bfs都试了一下，发现还是bfs稍微快一些，快了60ms。

搜索路径吧，然后比较和。

Java AC BFS

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
			return false;
		}
        return (bfs(root, sum));
    }
    public boolean bfs(TreeNode root, int sum){
        Queue<NodeSum> nodeQueue = new LinkedList<NodeSum>();
        nodeQueue.offer(new NodeSum(root,root.val));
        while(!nodeQueue.isEmpty()){
            NodeSum node = nodeQueue.peek();
            nodeQueue.poll();
            if(node.node.left == null && node.node.right == null && node.sum == sum){
                return true;
            }
            TreeNode point = node.node;
            int newSum = node.sum;
            if(point.left != null){
                point = point.left;
                nodeQueue.offer(new NodeSum(point,(newSum + point.val)));
            }
            point = node.node;
            if(point.right != null){
                point = point.right;
                nodeQueue.offer(new NodeSum(point,(newSum + point.val)));
            }
        }
        return false;
    }
    
    public class NodeSum {
        int sum ;
        TreeNode node;
        NodeSum(TreeNode root ,int sum){
            super();
            this.node = root;
            this.sum = sum;
        }
    }
}

Java AC DFS

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public boolean hasPathSum(TreeNode root, int sum) {
        if (root == null) {
			return false;
		}
        return (dfs(root, sum , root.val));
    }
    public boolean dfs(TreeNode root, int sum , int allSum){
        if (root.left == null && root.right == null ) {
			if (allSum == sum) {
				return true;
			}
		}
    	TreeNode point = root;
    	if (point.left != null) {
    		point = point.left;
			if (dfs(point, sum, allSum + point.val)) {
				return true;
			}
		}
    	point = root;
    	if (point.right != null) {
    		point = point.right;
    		if (dfs(point, sum, allSum + point.val)) {
				return true;
			}
		}
    	return false;
    }
    
    public class NodeSum {
        int sum ;
        TreeNode node;
        NodeSum(TreeNode root ,int sum){
            super();
            this.node = root;
            this.sum = sum;
        }
    }
}