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Binary Tree Longest Consecutive Sequence

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摘要:題目鏈接這一個(gè)類型的題都一樣用,分治的思想。兩種方式一種用,另一種直接把的長(zhǎng)度作為返回值,思路都一樣。也可以解,用或者來做,但是本質(zhì)都是。用用返回值在當(dāng)前層處理分別處理左右節(jié)點(diǎn),這樣不用傳上一次的值,注意這樣初始的就是了

Binary Tree Longest Consecutive Sequence

題目鏈接:https://leetcode.com/problems...

這一個(gè)類型的題都一樣用dfs,分治的思想。兩種方式:一種用global variable,另一種直接把sequence的長(zhǎng)度作為返回值,思路都一樣。也可以直接在當(dāng)前層對(duì)左右節(jié)點(diǎn)分別處理,本質(zhì)和前面一樣的。iteration也可以解,用stack或者queue來做,但是本質(zhì)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.在當(dāng)前層處理分別處理左右節(jié)點(diǎn),這樣不用傳上一次的值,注意這樣初始的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);
        }
    }

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