如何从数组表示构建不完整的二叉树

Question

如果输入是一个数组，其中null表示没有节点。

输入：

[1, 2, 3, null, 5, null, 7]

请假设我已经检查了输入。

对于每个array[i]，它的父array[i / 2]不会是null（递归，所以根不能是null）。

如何用这样的逻辑关系构建一棵树：

每个节点应该由一个 TreeNode 对象表示：

class TreeNode {
public:
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
};

我找到了一个 blog here 在那里建造了一棵完整的树

但是如果树像上面说的那样不完整，如何整齐高效地做呢？

测试数据：

[输入数组]

[-64,12,18,-4,-53,null,76,null,-51,null,null,-93,3,null,-31,47,null,3,53,-81,33,4,null,-51,-44,-60,11,null,null,null,null,78,null,-35,-64,26,-81,-31,27,60,74,null,null,8,-38,47,12,-24,null,-59,-49,-11,-51,67,null,null,null,null,null,null,null,-67,null,-37,-19,10,-55,72,null,null,null,-70,17,-4,null,null,null,null,null,null,null,3,80,44,-88,-91,null,48,-90,-30,null,null,90,-34,37,null,null,73,-38,-31,-85,-31,-96,null,null,-18,67,34,72,null,-17,-77,null,56,-65,-88,-53,null,null,null,-33,86,null,81,-42,null,null,98,-40,70,-26,24,null,null,null,null,92,72,-27,null,null,null,null,null,null,-67,null,null,null,null,null,null,null,-54,-66,-36,null,-72,null,null,43,null,null,null,-92,-1,-98,null,null,null,null,null,null,null,39,-84,null,null,null,null,null,null,null,null,null,null,null,null,null,-93,null,null,null,98]

Answer 1

当将二叉树实现为数组时，有一个两种表示法如何反映一个的清晰可视化另一个，并复习下划线的数学结构关系。

如果我们考虑索引为 0 的数组，数学关系可以被分解成这样，

根节点的索引为 0

对于i:th节点（i是数组索引）我们有（验证）

节点的left-child有索引2i + 1
节点的right-child有索引2(i + 1)
节点的parent有索引floor((i-1)/2)

所以，对于二叉树

如果我们让-表示null，则表示为

[0:a, 1:b, 2:c, 3:d, 4:e, 5:-, 6:-, 7:-, 8:-, 9:g, 10:-, 11:-, 12:-, 13:-, 14:-]

因此，现在要从数组创建 OO 表示，您只需应用这些索引规则。所以，既然你知道根节点是 a 那么我们就可以得到它的 children at:

左：2*0 + 1 = 1 => b
右：2*(0 + 1) = 2 => c

伪代码

for (int idx = 0; 2*(idx + 1) < len(arr); idx++) {
    if (arr[idx] == null) {
        // There is no node to add for this index
        continue;
    }

    TreeNode t = null;

    if (idx == 0) {
        // Root node case
        t = TreeNode(val: arr[idx]);
        binary_tree.add(id: idx, node: t);
    }

    // We do not know if these exist yet
    int left_idx = 2*idx + 1; 
    int right_idx = 2*(idx + 1);

    if (left_idx >= len(arr)) {
        // left_idx is out of bounds with respect to the array, 
        // and by extension so would the right node be
        continue;
    }

    TreeNode left = null;
    TreeNode right = null;

    if (arr[left_idx] != null) {
        // This node has never been encountered before
        // and it is non-null so it must be created.
        //
        // Since we know we have a root node then there is
        // no need to check if the tree already contains this
        // node, it simply is not possible. Ditto for the right
        // node.
        left = TreeNode(val: arr[left_idx]);
        binary_tree.add(id: left_idx, node: left);
    }

    if (right_idx >= len(arr)) {
        // There cannot be a right child
        continue;
    }

    if (arr[right_idx] != null) {
        // This node has never been encountered before
        // and it is non-null so it must be created.
        right = TreeNode(val: arr[right_idx]);
        binary_tree.add(id: right_idx, right);
    }

    // It does not matter if left or right is null
    t.set_left(left)
    t.set_right(right)    
}

Answer 2

就用递归遍历使用数组索引遍历节点，用Integer允许null。

private TreeNode array2Tree(Integer[] data,TreeNode root, int index){

    if(index >= data.length){
      return root;
    }

    if(data[index] != null){
      TreeNode temp =  new TreeNode(data[index]);
      root = temp;
      root.left = array2Tree(data,root.left,2*index+1);
      root.right = array2Tree(data,root.right,2*index+2);
    }

    return root;
}

Answer 3

我想这个例子可以解释你的想法。

array : [5,4,8,11,null,17,4,7,null,null,null,5]
Tree : 

                      5
                     /  \
                    4    8
                   /    / \
                  11   17  4
                 /        /
                7        5

以上所有答案都将输入数组视为一棵完整的树。所以 left.child=2idx+1 , right.child = 2idx+2 但实际上是错误的。因为那些

[5,4,8,11,null,17,4,7,null,null,null,5]
[5,4,8,11,null,17,4,7,null,null,null,null,null,5,null]

不同

这是我的解决方案

public static TreeNode createTree(Integer[] array) {
    if (array == null || array.length==0) {
        return null;
    }

    Queue<TreeNode> treeNodeQueue = new LinkedList<>();
    Queue<Integer> integerQueue = new LinkedList<>();
    for (int i = 1; i < array.length; i++) {
        integerQueue.offer(array[i]);
    }

    TreeNode treeNode = new TreeNode(array[0]);
    treeNodeQueue.offer(treeNode);

    while (!integerQueue.isEmpty()){
        Integer leftVal = integerQueue.isEmpty() ? null : integerQueue.poll();
        Integer rightVal = integerQueue.isEmpty() ? null : integerQueue.poll();
        TreeNode current = treeNodeQueue.poll();
        if (leftVal !=null) {
                TreeNode left = new TreeNode(leftVal);
                current.left = left;
                treeNodeQueue.offer(left);
        }
        if (rightVal !=null){
                TreeNode right = new TreeNode(rightVal);
                current.right = right;
                treeNodeQueue.offer(right);
        }
    }
    return treeNode;
}

Answer 4

在Java中：

public class TreeNode {
    public int val;
    public TreeNode left;
    public TreeNode right;
    public TreeNode() {}
    public TreeNode(int val) { this.val = val; }
    public TreeNode(int val, TreeNode left, TreeNode right) {
        this.val = val;
        this.left = left;
        this.right = right;
    }

    /**
     * Create a tree from array using levels i.e {-2,3,4,null,null,5,null,null,null,null,null, 6 } becomes 2 le-> 3, 2 re->4, 4 le->5, 5 le->6
     * @param arr the arr to be converted to a tree
     * @return
     */
    public static TreeNode createTreeFromArray(Integer[] arr){
        TreeNode root = new TreeNode();
        return  insertLevelOrder(arr, root, 0);
    }


    static TreeNode insertLevelOrder(Integer[] arr, TreeNode root,
                                 int i)
    {

        // Base case for recursion
        if (i < arr.length) {
            if(arr[i] == null)
                return null;
            TreeNode temp = new TreeNode(arr[i]);
            root = temp;

            // insert left child
            root.left = insertLevelOrder(arr, root.left,
                     2* i + 1);

            // insert right child
            root.right = insertLevelOrder(arr, root.right,
                     2* i + 2);
        }
        return root;
    }
}

Answer 5

谢谢史蒂文。我将 Steven 的 Java 代码转换为 Python。它对我有用！

class TreeNode:
    def __init__(self, val=0, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right

def creatBTree(data):
    if data == None or len(data) == 0:
        return None

    treeNodeQueue = []
    integerQueue = []

    for i in range(1,len(data)):
        print(i)
        integerQueue.append(data[i])

    treeNode = TreeNode(data[0])
    treeNodeQueue.append(treeNode)

    while integerQueue:
        if integerQueue:
            leftVal = integerQueue.pop(0)
        if integerQueue:
            rightVal = integerQueue.pop(0)

        current = treeNodeQueue.pop(0)

        if leftVal is not None:
            left = TreeNode(leftVal)
            current.left = left
            treeNodeQueue.append(left)
        if rightVal is not None:
            right = TreeNode(rightVal)
            current.right = right
            treeNodeQueue.append(right)

    return treeNode

如何从数组表示构建不完整的二叉树

How to build an incomplete binary tree from array representation

binary-tree

data-structures

伪代码