111.minimum depth of binary tree

题目链接：https://leetcode.com/problems/minimum-depth-of-binary-tree/description/

题目大意：求解二叉树的最小高度。

法一：BFS。新写一个class，承接TreeNode和当前结点高度。代码如下（耗时6ms）：

    static class MergeTree {
        TreeNode root;
        int depth;
        MergeTree(TreeNode r, int d) {
            root = r;
            depth = d;
        }
    }
    public int minDepth(TreeNode root) {
        int res = Integer.MAX_VALUE;
        if(root == null) {
            return 0;
        }
        Queue<MergeTree> q = new LinkedList<MergeTree>();
        MergeTree mT = new MergeTree(root, 1);
        q.offer(mT);
        while(!q.isEmpty()) {
            MergeTree tmpT = q.poll();
            TreeNode tmpN = tmpT.root;
            if(tmpN.left != null) {
                mT = new MergeTree(tmpN.left, tmpT.depth + 1);
                q.offer(mT);
            }
            if(tmpN.right != null) {
                mT = new MergeTree(tmpN.right, tmpT.depth + 1);
                q.offer(mT);
            }
            if(tmpN.left == null && tmpN.right == null) {
                res = Math.min(res, tmpT.depth);
            }
        }
        return res;
    }

同BFS，这里用两个queue，一个queue承接结点，一个queue承接当前结点的高度，免去新建一个class。代码如下（耗时8ms）：

    public int minDepth(TreeNode root) {
        int res = Integer.MAX_VALUE;
        if(root == null) {
            return 0;
        }
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        Queue<Integer> de = new LinkedList<Integer>();
        q.offer(root);
        de.offer(1);
        while(!q.isEmpty()) {
            TreeNode tmp = q.poll();
            int t = de.poll();
            if(tmp.left != null) {
                q.offer(tmp.left);
                de.offer(t + 1);
            }
            if(tmp.right != null) {
                q.offer(tmp.right);
                de.offer(t + 1);
            }
            if(tmp.left == null && tmp.right == null) {
                res = Math.min(res, t);
            }
        }
        return res;
    }

法二：DFS。竟然这个快这么多。代码如下（耗时1ms）：

    public int minDepth(TreeNode root) {
        if(root == null) {
            return 0;
        }
        return dfs(root, Integer.MAX_VALUE, 1);
    }
    private static int dfs(TreeNode root, int res, int cnt) {
        if(root.left == null && root.right == null) {
            return Math.min(res, cnt);
        }
        if(root.left != null) {
            res = dfs(root.left, res, cnt + 1);
        }
        if(root.right != null) {
            res = dfs(root.right, res, cnt + 1);
        }
        return res;
    }