已知二叉树前序，中序遍历，求后序遍历，java实现

  简单介绍一下思想，先看前序，前序遍历的第一个节点，就是该树的根。在中序中找到该根的位置，设为index，在中序遍历集合中，位于index之前的属于根的左子树，位于index之后的属于根的右子树。然后，对左右子数，遍历的调用该过程，就能将树结构还原。之后后序遍历也就不是问题了。

public class M {
	public static class Tree{
		public int value;
		public Tree left;
		public Tree right;
		public Tree(int value) {
			this.value=value;
		}
	}
	public static void main(String[] args){
		int[]preOrder={1,2,4,5,3,6,7};
		int[]midOrder={4,2,5,1,6,3,7};
		getBehindOrder(preOrder, midOrder);
	}
	public static void getBehindOrder(int[]pre,int[]mid){
		if (pre==null||mid==null||pre.length!=mid.length) {
			return;
		}
		Tree root=buildTree(pre, mid);
		behindOrder(root);
	}
	
	public static void behindOrder(Tree root){//后序遍历二叉树
		if (root==null) {
			return;
		}
		if (root.left!=null) {
			behindOrder(root.left);
		}
		if (root.right!=null) {
			behindOrder(root.right);
		}
		System.out.print(root.value+" ");
		
	}
	public static Tree buildTree(int[]preOrder,int[]midOrder){//根据前序和中序，建立二叉树
		int value=preOrder[0];
		int length=preOrder.length;
		Tree root=new Tree(value);
		root.left=root.right=null;
		if (preOrder.length==1) {
			return root;
		}
		int index=0;
		while(midOrder[index]!=value)
			index++;//此处还要考虑index==length-1的情况
		if (index>0) {
			//中序中，根节点左边的节点都属于左子树
			int[]leftSubPreOrder=new int[index];
			for(int i=0;i<leftSubPreOrder.length;i++){
				leftSubPreOrder[i]=preOrder[i+1];
			}
			int[]leftSubMidOrder=new int[index];
			for(int i=0;i<leftSubMidOrder.length;i++){
				leftSubMidOrder[i]=midOrder[i];
			}
			root.left=buildTree(leftSubPreOrder, leftSubMidOrder);
		}
		
		
		if(length-index-1>0){
		int[]rightSubMidOrder=new int[length-index-1];
		for(int i=0;i<rightSubMidOrder.length;i++){
			rightSubMidOrder[i]=midOrder[i+index+1];
		}
		int[]rightSubPreOrder=new int[length-index-1];
		
		for(int i=0;i<rightSubPreOrder.length;i++){
			rightSubPreOrder[i]=preOrder[i+index+1];
		}
		
		root.right=buildTree(rightSubPreOrder, rightSubMidOrder);
		}
		return root;
	}
}