题114 二叉树展开为链表

1.题目

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2.自己初解的思路及问题

略,直接看力扣题解

3.题解方法一:前序遍历(递归+迭代)

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递归:

class Solution {
    
    
    public void flatten(TreeNode root) {
    
    
        List<TreeNode> list = new ArrayList<TreeNode>();
        preorderTraversal(root, list);
        int size = list.size();
        for (int i = 1; i < size; i++) {
    
    
            TreeNode prev = list.get(i - 1), curr = list.get(i);
            prev.left = null;
            prev.right = curr;
        }
    }

    public void preorderTraversal(TreeNode root, List<TreeNode> list) {
    
    
        if (root != null) {
    
    
            list.add(root);
            preorderTraversal(root.left, list);
            preorderTraversal(root.right, list);
        }
    }
}

以下代码通过迭代实现前序遍历。

class Solution {
    
    
    public void flatten(TreeNode root) {
    
    
        List<TreeNode> list = new ArrayList<TreeNode>();
        Deque<TreeNode> stack = new LinkedList<TreeNode>();
        TreeNode node = root;
        while (node != null || !stack.isEmpty()) {
    
    
            while (node != null) {
    
    
                list.add(node);
                stack.push(node);
                node = node.left;
            }
            node = stack.pop();
            node = node.right;
        }
        int size = list.size();
        for (int i = 1; i < size; i++) {
    
    
            TreeNode prev = list.get(i - 1), curr = list.get(i);
            prev.left = null;
            prev.right = curr;
        }
    }
}

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4.题解方法二:前序遍历和展开同步进行

4.题解方法三:寻找前驱节点

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转载自blog.csdn.net/ambitionLlll/article/details/114365490