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  • 105. Construct Binary Tree from Inorder and preorder Traversal

        /*
         * 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; }
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  • 原文地址:https://www.cnblogs.com/zmyvszk/p/5503911.html
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