生成二叉树

1.Construct Binary Tree from Inorder and Postorder Traversal

1. 问题描述
   由中序遍历和后续遍历生成二叉树
2. 解题思路
   方法一：时间O(n2) ，空间 O(1)

public TreeNode buildTree(int[] inorder, int[] postorder) {
        if(inorder == null || postorder == null)
            return null;
        int iLength = inorder.length;
        int pLength = postorder.length;
        if(iLength == 0 || pLength == 0)
            return null;
        return buildNode(inorder,0,iLength-1,postorder,0,pLength-1);
    }

    public TreeNode buildNode(int[] inorder, int iB, int iE, int[] postorder, int pB, int pE) {
        if(iB == iE){
            TreeNode node = new TreeNode(inorder[iB]);
            return node;
        }else if(iB > iE)
            return null;
        TreeNode root = new TreeNode(postorder[pE]);
        int index = findIndex(inorder,iB,iE,postorder[pE]);
        int llength = index - iB;
        int leftBeginIn = iB;
        int leftEndIn = index - 1;
        int leftBeginPost = pB;
        int leftEndPost = pB +  llength - 1;
        root.left = buildNode(inorder,leftBeginIn,leftEndIn,postorder,leftBeginPost,leftEndPost);
        int rightBeginIn = index + 1;
        int rightEndIn = iE;
        int rightBeginPost = pB +  llength;
        int rightEndPost = pE - 1;
        root.right = buildNode(inorder,rightBeginIn,rightEndIn,postorder,rightBeginPost,rightEndPost);
        return root;

    }

    public int findIndex(int[] Nums,int iB,int iE,int target){
        for(int i = iB; i <= iE; i++){
            if(Nums[i] == target)
                return i;
        }
        return -1;
    }

方法二：使用哈希表。时间复杂度O(n)，空间O(n)

public TreeNode buildTreePostIn(int[] inorder, int[] postorder) {
    if (inorder == null || postorder == null || inorder.length != postorder.length)
        return null;
    HashMap hm = new HashMap();
    for (int i=0;ireturn buildTreePostIn(inorder, 0, inorder.length-1, postorder, 0, 
                          postorder.length-1,hm);
}

private TreeNode buildTreePostIn(int[] inorder, int is, int ie, int[] postorder, int ps, int pe, 
                                 HashMap hm){
    if (ps>pe || is>ie) return null;
    TreeNode root = new TreeNode(postorder[pe]);
    int ri = hm.get(postorder[pe]);
    TreeNode leftchild = buildTreePostIn(inorder, is, ri-1, postorder, ps, ps+ri-is-1, hm);
    TreeNode rightchild = buildTreePostIn(inorder,ri+1, ie, postorder, ps+ri-is, pe-1, hm);
    root.left = leftchild;
    root.right = rightchild;
    return root;
}

2.Construct Binary Tree from Preorder and Inorder Traversal

1. 问题描述
   由先序遍历和中续遍历生成二叉树
2. 解题思路
   方法一：时间O(n) ，空间 O(n)

public TreeNode buildTree(int[] preorder, int[] inorder) {
        if(preorder == null || inorder == null)
            return null;
        int preLength = preorder.length;
        int inLength = inorder.length;
        HashMap map = new HashMap<>();
        for(int i = 0; i < inLength; i++){
            map.put(inorder[i], i);
        }
        return buildNode(preorder,0,preLength-1,inorder,0,inLength-1,map);
    }

public TreeNode buildNode(int[] preorder, int pB, int pE, int[] inorder, int iB, int iE,HashMap map){
        if(iB > iE)
            return null;
        if(iB == iE){
            TreeNode root = new TreeNode(inorder[iB]);
            return root;
        }
        TreeNode root = new TreeNode(preorder[pB]);
        int index = map.get(preorder[pB]);
        int length = index - iB;
        root.left = buildNode(preorder, pB+1, pB+length, inorder, iB, index-1,map);
        root.right = buildNode(preorder, pB+length+1, pE, inorder, index+1, iE,map);
        return root;
    }