Construct Binary Tree from Inorder and Postorder Traversal

标签：重构   binary   inorder   postorder   

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:
You may assume that duplicates do not exist in the tree.

还是重构树  与上一题相比  先序换成了后序  思路还是一样 代码如下：

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(postorder.length==0&&inorder.length==0){
			return null;
		}
		TreeNode root=new TreeNode(postorder[postorder.length-1]);
		int begindex=0;
		for(int i=0;i<inorder.length;i++){
			if(inorder[i]==postorder[postorder.length-1]){
				begindex=i;
				break;
			}
		}
		int leftlen=begindex;
	    int rightlen=inorder.length-begindex-1;
	    int[] leftpost=new int[leftlen];
	    int[] leftino=new int[leftlen];
		for(int i=0;i<begindex;i++){
			leftpost[i]=postorder[i];
			leftino[i]=inorder[i];			
		}
		root.left=buildTree(leftino, leftpost);
		int[] rightpost=new int[rightlen];
		int[] rightino=new int[rightlen];
		for(int i=0;i<rightlen;i++){
			rightpost[i]=postorder[i+begindex];
			rightino[i]=inorder[i+begindex+1];			
		}
		root.right=buildTree(rightino, rightpost);
		return root;
    }
}

Construct Binary Tree from Inorder and Postorder Traversal

标签：重构   binary   inorder   postorder   

原文地址：http://blog.csdn.net/u012734829/article/details/42265519