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  • 【leetcode】366.Find Leaves of Binary Tree

    原题

    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].

    Explanation:

    1. Removing the leaves [4, 5, 3] would result in this tree:
      1
      /
      2
    2. Now removing the leaf [2] would result in this tree:
      1
    3. Now removing the leaf [1] would result in the empty tree:
      []
      Returns [4, 5, 3], [2], [1].

    解析

    给一颗二叉树,
    返回这个二叉树的叶子节点,每一次一组,依次放在List中

    思路

    我的思路:每一次深度遍历收割一次叶子放在结果中(缺点:深度有多少,就要深度遍历多少次)
    优化算法:深度优选搜索,叶子节点深度标记为0,其他节点的深度标记为左右叶子节点较大的深度+1;相同深度的放在一个结果节点中

    我的解法

    public List<List<Integer>> FindLeaves(TreeNode root) {
            List<List<Integer>> result = new ArrayList<>();
            while (root != null) {
                List<Integer> re = new ArrayList<>();
                if (DFSTree(root, re)) {
                    root = null;
                }
                result.add(re);
            }
            return result;
        }
    
        private boolean DFSTree(TreeNode root, List<Integer> re) {
            if (root.left == null && root.right == null) {
                re.add(root.val);
                return true;
            }
            if (root.left != null) {
                if (DFSTree(root.left, re)) {
                    root.left = null;
                }
            }
            if (root.right != null) {
                if (DFSTree(root.right, re)) {
                    root.right = null;
                }
            }
            return false;
        }
    

    最优解

        public List<List<Integer>> FindLeavesOptimize(TreeNode root) {
            List<List<Integer>> result = new ArrayList<>();
            DFSTreeOptimize(root, result);
            return result;
        }
        private int DFSTreeOptimize(TreeNode root, List<List<Integer>> result) {
            if (root.left == null && root.right == null) {
                if (result.size() <= 0) {
                    result.add(0, new ArrayList<>());
                }
                result.get(0).add(root.val);
                return 0;
            }
            int deep, leftDeep = 0, rightDeep = 0;
            if (root.left != null) {
                leftDeep = DFSTreeOptimize(root.left, result);
            }
            if (root.right != null) {
                rightDeep = DFSTreeOptimize(root.right, result);
            }
            deep = Math.max(leftDeep, rightDeep) + 1;
    
            if (result.size() <= deep) {
                result.add(deep, new ArrayList<>());
            }
            result.get(deep).add(root.val);
            return deep;
        }
    
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  • 原文地址:https://www.cnblogs.com/shanelau/p/7238089.html
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