Binary Tree Longest Consecutive Sequence

Binary Tree Longest Consecutive Sequence

题目链接：https://leetcode.com/problems...

这一个类型的题都一样用dfs，分治的思想。两种方式：一种用global variable，另一种直接把sequence的长度作为返回值，思路都一样。也可以直接在当前层对左右节点分别处理，本质和前面一样的。iteration也可以解，用stack或者queue来做，但是本质都是dfs。
1.用global variable

    public int longestConsecutive(TreeNode root) {
        /* dfs
         * arguments: curNode, previous value, length
         * 2 chooses: 1. curNode.val >= previous_value => length + 1
         *            2. curNode.val < previous_value  => length = 1
         * update length each recursion: use a global variable
         */
         if(root == null) return 0;
         dfs(root, root.val - 1, 0);
         return global;
    }
    
    int global = 0;
    private void dfs(TreeNode curNode, int previous_value, int len) {
        // update global length
        global = Math.max(global, len);
        // base case
        if(curNode == null) {
            return;
        }
        
        if(curNode.val - previous_value == 1) len++;
        else len = 1;
        dfs(curNode.left, curNode.val, len);
        dfs(curNode.right, curNode.val, len);
    }

2.用返回值

    public int longestConsecutive(TreeNode root) {
        /* dfs
         * arguments: curNode, previous value, length
         * 2 chooses: 1. curNode.val >= previous_value => length + 1
         *            2. curNode.val < previous_value  => length = 1
         * update length each recursion: return the max length
         */
         if(root == null) return 0;
         return dfs(root, root.val - 1, 0);
    }
    
    private int dfs(TreeNode curNode, int previous_value, int len) {
        // base case
        if(curNode == null) {
            return len;
        }
        
        if(curNode.val - previous_value == 1) len++;
        else len = 1;
        return Math.max(len, Math.max(dfs(curNode.left, curNode.val, len), dfs(curNode.right, curNode.val, len)));
    }

3.在当前层处理分别处理左右节点，这样不用传上一次的值，注意这样初始的len就是1了：

    public int longestConsecutive(TreeNode root) {
         if(root == null) return 0;
         dfs(root, 1);
         return global;
    }
    
    int global = 0;
    private void dfs(TreeNode curNode, int len) {
        global = Math.max(global, len);
        
        if(curNode.left != null) {
            if(curNode.val + 1 == curNode.left.val) dfs(curNode.left, len+1);
            else dfs(curNode.left, 1);
        }
        if(curNode.right != null) {
            if(curNode.val + 1 == curNode.right.val) dfs(curNode.right, len+1);
            else dfs(curNode.right, 1);
        }
    }