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  • linked list焦点问题,面经里很多,考虑相交不相交,有环无环 + Find Leaves of Binary Tree (Java)

    break the loop at the last node which pointed to the entry.

    Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

    Example:
    Given binary tree

              1
             / 
            2   3
           /      
          4   5    
    

    Returns [4, 5, 3], [2], [1].

    ollowup: 每个node还是有left,right指针,但是结构并非tree,而是graph怎么算, 考虑有/无circular dependency 两种情况;
    方法是用hashmap 记录计算过的node level 和 一个Set 记录dependent node
     

    For this question we need to take bottom-up approach. The key is to find the height of each node. Here the definition of height is:
    The height of a node is the number of edges from the node to the deepest leaf. 

    I used a helper function to return the height of current node. According to the definition, the height of leaf is 0. h(node) = 1 + max(h(node.left), h(node.right)).
    The height of a node is also the its index in the result list (res). For example, leaves, whose heights are 0, are stored in res[0]. Once we find the height of a node, we can put it directly into the result.

    public List<List<Integer>> findLeaves(TreeNode root) {
        List<List<Integer>> result = new ArrayList<List<Integer>>();
        helper(result, root);
        return result;
    }
     
    // traverse the tree bottom-up recursively
    private int helper(List<List<Integer>> list, TreeNode root){
        if(root==null)
            return -1;
     
        int left = helper(list, root.left);
        int right = helper(list, root.right);
        int curr = Math.max(left, right)+1;
     
        // the first time this code is reached is when curr==0,
        //since the tree is bottom-up processed.
        if(list.size()<=curr){
            list.add(new ArrayList<Integer>());
        }
     
        list.get(curr).add(root.val);
     
        return curr;
    }
    

      

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  • 原文地址:https://www.cnblogs.com/apanda009/p/7952951.html
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