Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2024-01-17 14:50:43 发布

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最新推荐文章于 2024-01-17 14:50:43 发布

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分类专栏： LeetCode

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本文介绍两种构建二叉树的方法：递归法与迭代法。递归方法通过不断缩小子问题规模直至基本情况来构造树结构；迭代方法则使用栈辅助实现，避免了递归带来的栈溢出风险。

    //recursive
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(preorder.length == 0 || preorder.length != inorder.length) return null;
        return buildTree(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1);
    }
    
    public TreeNode buildTree(int[] preorder, int[] inorder, int prestart, int preend, int instart, int inend) {
        if(prestart > preend || instart > inend) return null;
       TreeNode root = new TreeNode(preorder[prestart]);
       if(prestart == preend) return root;
       int index = instart;
       while(index <= inend && inorder[index] != preorder[prestart]) index++;
       root.left = buildTree(preorder, inorder, prestart + 1, index - instart + prestart, instart, index - 1);
       root.right = buildTree(preorder, inorder, index - instart + prestart + 1, preend, index + 1, inend);
       return root;
    }



    //iterative
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        // Start typing your Java solution below
        // DO NOT write main() function
        if(preorder.length == 0 || preorder.length != inorder.length) return null;
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode root = new TreeNode(preorder[0]);
        stack.push(root);
        int preindex = 1;
        int inindex = 0;
        TreeNode current = root;
        while(preindex < preorder.length) {
            boolean flag = false;
            while(!stack.empty() && inorder[inindex] == stack.peek().val) {
                current = stack.pop();
                inindex++;
                flag = true;
            }
            TreeNode next = new TreeNode(preorder[preindex++]);
            stack.push(next);
            if(flag == false) {
                current.left = next;
            }else {
                current.right = next;
            }
            current = next;
        }
        return root;
    }