106. Construct Binary Tree from Inorder and Postorder Traversal

和105一样的。。这个是给IN 和 POST

POST的最后一个是ROOT

所以就从后找，几乎一样的

 public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder.length == 0) return null;

        int length = postorder.length;
        int tailVal = postorder[length-1];
        int tempIndex = length-1;

        while(tailVal != inorder[tempIndex])
        {
        	tempIndex--;
        }

        TreeNode root = new TreeNode(tailVal);
        int[] newIn = Arrays.copyOfRange(inorder,tempIndex+1,length);
        int[] newPost = Arrays.copyOfRange(postorder,tempIndex,length-1);
        root.right = buildTree(newIn,newPost);

        newIn = Arrays.copyOfRange(inorder,0,tempIndex);
        newPost = Arrays.copyOfRange(postorder,0,tempIndex);
        root.left = buildTree(newIn,newPost);


        return root;

    }

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二刷。

二刷做的我想打人了。

尝试不建新Array，迭代4个index，结果没找对关系，浪费了好多时间。

最后老老实实重新copyRangeOf了。

感觉直接用INDEX是可以实现的。

public class Solution 
{
    public TreeNode buildTree(int[] in, int[] po) 
    {
        if(in.length == 0) return null;
        return generate(in,po);
    }
    
    public TreeNode generate(int[] in, int[] po)
    {
        if(in.length == 0) return null;
        if(in.length == 1) return new TreeNode(po[0]);
        
        int rootVal = po[po.length-1];
        int temp = 0;
        TreeNode root = new TreeNode(rootVal);
        for(int i = 0; i < in.length ;i++)
        {
            if(in[i] == rootVal)
            {
                temp = i;
                break;
            }
        }
        
      
            int[] leftIn = Arrays.copyOfRange(in,0,temp);
            int[] leftPo = Arrays.copyOfRange(po,0,temp);
            root.left = generate(leftIn,leftPo);            

            int[] rightIn = Arrays.copyOfRange(in,temp+1,in.length);
            int[] rightPo = Arrays.copyOfRange(po,temp,po.length-1);
    
            root.right = generate(rightIn,rightPo);
        

        
        return root;
    }
}

Got my ass kicked by 90%+ ppl.

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三刷。

postOrder的最后一个一定是当前ROOT。
在InOrder里寻找ROOT.VAL，找到位置的左边是LEFT，右边是RIGHT，进行递归。

一刷二刷是新建了ARRAY，没必要，直接记录进入递归的区间就可以了。

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return convert(inorder, 0, inorder.length-1, postorder, 0, postorder.length-1);
    }
    public TreeNode convert(int[] in, int inL, int inR, 
                            int[] post, int postL, int postR) {
        
        if (postL > postR) return null;
        if (postL == postR) return new TreeNode(post[postR]);
        
        TreeNode res = new TreeNode(post[postR]);
        int i = 0;
        while (post[postR] != in[inR - i]) i++;
        
        res.left = convert(in, inL, inR-i-1, post, postL, postR-i-1);
        res.right = convert(in, inR-i+1, inR, post, postR-i, postR-1);
        return res;
    }
}