leetcode——1028.从先序遍历还原二叉树

Posted 2020-12-11 欣姐姐

public TreeNode recoverFromPreorder(String S) {
        Stack<TreeNode> path = new Stack<>();  //构建好栈
        //定义一个变量来遍历S
        int i = 0;
        //定义一个int变量来确定节点的层数
        //先将S转换成char[]
        char[] s = S.toCharArray();
        //开始遍历S
        while(i<s.length){
            int level = 0;
            //先确定层数
            while(s[i] == \'-\'){
                ++i;
                ++level;
            }
            //再确定结点的值
            //先将值初始化为0
            int value = 0;
            while(i<s.length && Character.isDigit(s[i])){
                value = value * 10 + s[i] - \'0\';
                ++i;
            }
            //构造树的结点
            TreeNode node = new TreeNode(value);
            //插入到正确位置
            if(level == path.size()){
                if(!path.isEmpty()) {
                    path.peek().left = node;
                }
            }else{
                while (level < path.size()){
                    path.pop();
                }
                path.peek().right = node;
            }
            path.push(node);
        }
        while (path.size() > 1){
            path.pop();
        }
        return path.peek();
    }

 

 

只是看懂了官方题解又默写了一遍而已……太难了，我脑子跟不上……想哭………………

 

——2020.6.20