【LeetCode】Path Sum

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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;        }    }}

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