已知前序遍历序列和中序遍历序列，求二叉树的后序遍历

题目

已知前序遍历序列和中序遍历序列，求二叉树的后序遍历

思路：

在前序遍历的序列中第一个就是树的根结点，此时再在中序遍历的序列里查找这个根结点，则中序遍历的序列里根结点左侧的就是左子树，右侧的就是右子树，再对左右子树进行同样的操作，此时可以使用递归实现，这样便能构造出这个二叉树。

class TreeNode{

private int value;
private TreeNode left;
private TreeNode right;

public TreeNode(int value) {
super();
this.value = value;
}

public TreeNode(int value, TreeNode left, TreeNode right) {
super();
this.value = value;
this.left = left;
this.right = right;
}

public int getValue() {
return value;
}

public void setValue(int value) {
this.value = value;
}

public TreeNode1 getLeft() {
return left;
}

public void setLeft(TreeNode left) {
this.left = left;
}

public TreeNode1 getRight() {
return right;
}

public void setRight(TreeNode right) {
this.right = right;
}

}

public class Item {

public static void main(String[] args) {

// 二叉树的前序遍历
int preOrder[] = { 1, 2, 4, 7, 3, 5, 6, 8 };
// 二叉树的中序遍历
int inOrder[] = { 4, 7, 2, 1, 5, 3, 8, 6 };

TreeNode root = constructTree(preOrder, inOrder);
printPostOrder(root);

}

//重建二叉树
private static TreeNode1 constructTree(int preOrder[], int inOrder[]){

//根据前序遍历创建根节点
TreeNode root = new TreeNode(preOrder[0]);
root.setLeft(null);
root.setRight(null);

//左子树的个数
int leftNum = 0;

//在中序遍历中找到根节点并得出左子树的个数
for (int i = 0; i < inOrder.length; i++) {
if (inOrder[i] == root.getValue()) {
break;
}
leftNum++;
}

//右子树的个数
int rightNum = inOrder.length - leftNum - 1;

if (leftNum > 0) {
//构造左子树的前序遍历和中序遍历
int leftPreOrder[] = new int[leftNum];
int leftInOrder[] = new int[leftNum];

for (int i = 0; i < leftInOrder.length; i++) {
leftPreOrder[i] = preOrder[1 + i];
leftInOrder[i] = inOrder[i];
}

//递归构建左子树
TreeNode leftRoot = constructTree(leftPreOrder, leftInOrder);
root.setLeft(leftRoot);

}

if (rightNum > 0) {
//构造右子树的前序遍历和中序遍历
int rightPreOrder[] = new int[rightNum];
int rightInOrder[] = new int[rightNum];

for (int i = 0; i < rightInOrder.length; i++) {
rightPreOrder[i] = preOrder[leftNum + 1 + i];
rightInOrder[i] = preOrder[leftNum + 1 + i];
}

//递归构建右子树
TreeNode rightRoot = constructTree(rightPreOrder, rightInOrder);
root.setRight(rightRoot);
}
return root;
}

//二叉树的后序遍历
public static void printPostOrder(TreeNode root){

if (root != null) {
printPostOrder(root.getLeft());
printPostO
4000
rder(root.getRight());
System.out.printf(root.getValue() + "  ");
}
}
}

感谢