1. 设计一个算法将顺序表L中所有小于0的整数放前半部分，大于等于0的整数放在后半部分
2. 二叉树的删除

设计一个算法将顺序表L中所有小于0的整数放前半部分，大于等于0的整数放在后半部分

思路:从左侧找出>0的元素，从右侧找出<=0的元素,然后进行交换

static void Move(SqList &l)
{int i=0,j=l.length-1;while(i<j){//position from left to rightwhile(i<j && l.elem[i]<0)i++;//position from right to leftwhile(i<j && l.elem[j]>=0)j--;//swapif(i<j){int temp=l.elem[i];l.elem[i]=l.elem[j];l.elem[j]=temp;}}
}

二叉树的删除

情况1：删除没有子节点的节点

即删除6,7,8,9,2这些节点

1. 先找到要删除的节点,并记录该节点属于左节点还是右节点
2. 判断要删除的节点是否为没有子节点,如果确实没有的话,就将其父节点的其中的一个左右节点置空

执行第1步,查找节点如果current不为空则执行第2步

      Node current = root;Node parent = root;boolean isLeftChild = true;while(current.iData != key)        // search for node{parent = current;if(key < current.iData)         // go left?{isLeftChild = true;current = current.leftChild;}else                            // or go right?{isLeftChild = false;current = current.rightChild;}if(current == null)             // end of the line,return false;                // didn't find it}  // end while// found node to delete

执行第2步,判断左右节点均未空

// if no children, simply delete it
if(current.leftChild==null &&current.rightChild==null){if(current == root)             // if root,root = null;                 // tree is emptyelse if(isLeftChild)parent.leftChild = null;     // disconnectelse                            // from parentparent.rightChild = null;}

情况2：有一个子节点的节点

第1步如情况1的第1步相同，还是先找到要删除的节点.

1. 若该节点的右节点为空,则将左节点提升为父亲节点
2. 若该节点的左节点为空,则将右节点提升为父亲节点
3. 从以上两点可得知将剩余子节点提升为父亲节点
4. 根据isLeftChild标志判断删除的节点是否为左节点

执行步骤1和步骤4

if(current.rightChild==null)if(current == root)root = current.leftChild;else if(isLeftChild)parent.leftChild = current.leftChild;elseparent.rightChild = current.leftChild;

执行步骤2和步骤4

// if no left child, replace with right subtree
if(current.leftChild==null)if(current == root)root = current.rightChild;else if(isLeftChild)parent.leftChild = current.rightChild;elseparent.rightChild = current.rightChild;

情况3：删除有两个子节点的节点

铺垫:

1.查找二叉排序树的最大值和最小值

因为二叉树的的节点总是大于左节点,小于右节点的,所以顺着左节点总是能找到最小值，同理顺着右节点总是能找到最大值

public Node getMin()
{
Node current = root;
Node last;
while(current!=null)       {last=current;current = current.leftChild;     }
return last;
} public Node getMax()
{
Node current = root;
Node last;
while(current!=null)       {last=current;current = current.rightChild;     }
return last;
}

2.二叉排序树的中序遍历

如上图

二叉排序树的中序遍历的值是一个升序排列,以上结果为15,20,25,30,50,60,70

寻找具有有两个子节点的节点的替补节点

如要删除该节点，要么被代替的节点需要具备以下特征：

1. 比该节点的左节点大
2. 比该节点的右节点小

如果无法满足以上两点,情况将变的更加复杂.

如要删除节点20,那么节点30则是最佳替补(这里如果有个节点25则更加说明这个情况).
节点25比15大,比30小.

中序后继:由于25在20后面,则成25为节点20的后继.所以当遇到要删除有两个节点的节点时,首要目标就是找到该节点的后继节点,以上的铺垫1内容就是为这里准备的。查找后继节点规则：

1. 如果有左节点,则最深处左节点为后继节点
2. 如果没有左节点，则第一个右节点为后继节点(如50的后继节点为60)

以下代码体现了以上2个步骤

private Node getSuccessor(Node delNode){Node successorParent = delNode;Node successor = delNode;Node current = delNode.rightChild;   // go to right childwhile(current != null)               // until no more{                                 // left children,successorParent = successor;successor = current;current = current.leftChild;      // go to left child}return successor;}

删除中序后继节点

找到中序后继后，还要处理一些事情.来考虑一个中序后继的一些特点:

1. 一定没有左节点(如果有就不是后继了，可以继续往左节点找)
2. 但有可能会有右节点

所以要按照情况2只有右节点的原则处理该右继节点

链接中序后继的右节点

即要删除的节点的右节点变成了中序后继节点的右节点了

所以在getSuccessor方法中while循环结束后，还需要做以下处理

if(successor != delNode.rightChild)  // right child,{                                 // make connectionssuccessorParent.leftChild = successor.rightChild;successor.rightChild = delNode.rightChild;}

链接中序后继的左节点

Node successor = getSuccessor(current);// connect parent of current to successor instead
if(current == root)root = successor;
else if(isLeftChild)parent.leftChild = successor;
elseparent.rightChild = successor;// connect successor to current's left child
successor.leftChild = current.leftChild;

可以看到删除一个后继的动作相当的复杂

1. 改变了其右节点的父亲节点
2. 改变了其右节点
3. 改变其父亲节点
4. 改变了其左节点

完整删除代码示例(来自Java数据结构和算法)

   public boolean delete(int key) // delete node with given key{                           // (assumes non-empty list)Node current = root;Node parent = root;boolean isLeftChild = true;while(current.iData != key)        // search for node{parent = current;if(key < current.iData)         // go left?{isLeftChild = true;current = current.leftChild;}else                            // or go right?{isLeftChild = false;current = current.rightChild;}if(current == null)             // end of the line,return false;                // didn't find it}  // end while// found node to delete// if no children, simply delete itif(current.leftChild==null &&current.rightChild==null){if(current == root)             // if root,root = null;                 // tree is emptyelse if(isLeftChild)parent.leftChild = null;     // disconnectelse                            // from parentparent.rightChild = null;}// if no right child, replace with left subtreeelse if(current.rightChild==null)if(current == root)root = current.leftChild;else if(isLeftChild)parent.leftChild = current.leftChild;elseparent.rightChild = current.leftChild;// if no left child, replace with right subtreeelse if(current.leftChild==null)if(current == root)root = current.rightChild;else if(isLeftChild)parent.leftChild = current.rightChild;elseparent.rightChild = current.rightChild;else  // two children, so replace with inorder successor{// get successor of node to delete (current)Node successor = getSuccessor(current);// connect parent of current to successor insteadif(current == root)root = successor;else if(isLeftChild)parent.leftChild = successor;elseparent.rightChild = successor;// connect successor to current's left childsuccessor.leftChild = current.leftChild;}  // end else two children// (successor cannot have a left child)return true;                                // success}  // end delete()
// -------------------------------------------------------------// returns node with next-highest value after delNode// goes to right child, then right child's left descendentsprivate Node getSuccessor(Node delNode){Node successorParent = delNode;Node successor = delNode;Node current = delNode.rightChild;   // go to right childwhile(current != null)               // until no more{                                 // left children,successorParent = successor;successor = current;current = current.leftChild;      // go to left child}// if successor notif(successor != delNode.rightChild)  // right child,{                                 // make connectionssuccessorParent.leftChild = successor.rightChild;successor.rightChild = delNode.rightChild;}return successor;}