将二叉树转换为排序数组 [英] Turning a Binary Tree to a sorted array

问题描述

有没有一种方法可以将Binary转换为排序的数组，而不必遍历每个数组索引的树?

Is there a way to turn a Binary to a sorted array without having to traverse the tree for every array index?

Node root;
Node runner;
int current_smallest;

void findsmallest(Node root){
    //Pre-order traversal
    if(root == null){
        return;
    }else{
        runner = root;
        if(current_smallest == null){
            current_smallest = runner.element;
        }else{
            if(current_smallest > runner.element){
            current_smallest = runner.element;
            }
        }
        findsmallest(runner.left);
        findsmallest(runner.right);
    }   
}

void fill_array( int [] array ){

    for(int i =0; i < array.length(); i++){
        findsmallest(root);
        array[i] = current_smallest;
    }
} 

如您所见，如果树中有很多节点，这可能会花费很长时间. 顺便说一句，我忘了表明必须从一开始就遍历整个树才能获得数组的长度.

As you can see this can be take a long time if there are a lot of nodes in the tree. Btw, I forgot to show that the whole tree would have to be traversed at the start to get the length of the array.

推荐答案

是的，您可以这样做:运行 按顺序遍历树 ，保留数组的当前位置，并将节点的值存储在数组的当前位置.

Yes, you can do that: run an in-order traversal of the tree, keep the current position of the array, and store the node's value at the then-current position of the array.

您可以递归进行有序遍历，也可以使用堆栈数据结构进行遍历.如果要递归执行此操作，可以执行以下操作:

You can do in-order traversal recursively, or you can do it with a stack data structure. If you want to do it recursively, you can do this:

int fill_array(Node root, int [] array, int pos) {
    if (root.left != null) {
        pos = fill_array(root.left, array, pos);
    }
    array[pos++] = root.element;
    if (root.right != null) {
        pos = fill_array(root.right, array, pos);
    }
    return pos; // return the last position filled in by this invocation
}

请注意，为了使上述递归过程正常工作，调用者必须在传递给函数的array中分配足够的空间.

Note that in order for the above recursive procedure to work, the caller must allocate enough space in the array passed into the function.

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