105. Construct Binary Tree from Inorder and preorder Traversal

Posted 2020-07-07 我的名字叫周周

    /*
     * 105. Construct Binary Tree from Inorder and preorder Traversal 
     * 11.20 By Mingyang 
     * 千万不要以为root一定在中间那个点，还是要找一下中间点的位置
     * p.left = construct(preorder, preStart + 1, preStart + (k - inStart),inorder, inStart, k - 1);
     * p.right = construct(preorder, preStart + (k - inStart) + 1, preEnd,inorder, k + 1, inEnd);
     * 难点就在于如何找到这两个起点和终点的index
     */
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        int preEnd = preorder.length - 1;
        int inEnd = inorder.length - 1;
        return construct(preorder, 0, preEnd, inorder, 0, inEnd);
    }
    public TreeNode construct(int[] preorder, int preStart, int preEnd,int[] inorder, int inStart, int inEnd) {
        if (preStart > preEnd || inStart > inEnd) {
            return null;
        }
        int val = preorder[preStart];
        TreeNode p = new TreeNode(val);
        // find parent element index from inorder
        int k = 0;
        for (int i = 0; i < inorder.length; i++) {
            if (val == inorder[i]) {
                k = i;
                break;
            }
        }
        p.left = construct(preorder, preStart + 1, preStart + (k - inStart),inorder, inStart, k - 1);
        p.right = construct(preorder, preStart + (k - inStart) + 1, preEnd,inorder, k + 1, inEnd);
        return p;
    }