[Algorithm] Serialize and Deserialize Binary Tree

最新推荐文章于 2021-02-16 03:26:06 发布

转载最新推荐文章于 2021-02-16 03:26:06 发布 · 128 阅读

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原文链接：http://www.cnblogs.com/Answer1215/p/10487181.html

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#开发工具

本文介绍了一种实现二叉树序列化和反序列化的算法，使用递归方式将二叉树转换为字符串，并能从字符串中恢复原始二叉树结构。通过实例演示了如何构建二叉树，对其进行序列化，以及如何从序列化后的字符串进行反序列化。

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Given the root to a binary tree, implement serialize(root), which serializes the tree into a string, and deserialize(s), which deserializes the string back into the tree.

For example, given the following Node class
class Node:
    def __init__(self, val, left=None, right=None):
        self.val = val
        self.left = left
        self.right = right
The following test should pass:
node = Node('root', Node('left', Node('left.left')), Node('right'))
assert deserialize(serialize(node)).left.left.val == 'left.left'

Define createNode function:

function createNode(val, left = null, right = null) {
  return {
    val,
    left,
    addLeft(leftKey) {
      return this.left = leftKey ? createNode(leftKey) : null;
    },
    right,
    addRight(rightKey) {
      return this.right = rightKey ? createNode(rightKey) : null;
    }
  };
}

Define a binary tree:

function createBT(rootKey) {
  const root = createNode(rootKey);
  return {
    root,
    nodes: [],
    serialize(node) {

    },
    deserialize(str) {

    }
  };
}

Construct the tree:

const tree = createBT("root");
const root = tree.root;
const left = root.addLeft("left");
root.addRight("right");
left.addLeft("left.left");
left.addRight("left.right");

Serialize the tree:

The idea to serialize a tree is using recsurice apporach, keep calling serialize function for the left side node, then keep calling serialize function for right side nodes.

In the end, we output the serialized nodes in string format.

    serialize(node) {
      if (node) {
        this.nodes.push(node.val);
        this.serialize(node.left);
        this.serialize(node.right);
      } else {
        this.nodes.push("#");
      }
      return this.nodes.join(" ");
    },

const ser = tree.serialize(root); // root left left.left # # # right # #

Deserialize tree:

By given the serialized tree, every time we calling recsurice, we are trying to create Node, then append its left and right node.

    deserialize(str) {
      
      function* getVals() {
        for (let val of str.split(" ")) {
          yield val;
        }
      }
      
      const helper = (pointer) => {
        let { value , done } = pointer.next();
        if (value === "#" || done) return null;

        let node = createNode(value);
        node.left = helper(pointer);
        node.right = helper(pointer);
        return node;
      };
      return helper(getVals());
    }

const dec = tree.deserialize(ser).left.left.val; // left.left

-----

function createNode(val, left = null, right = null) {
  return {
    val,
    left,
    addLeft(leftKey) {
      return this.left = leftKey ? createNode(leftKey) : null;
    },
    right,
    addRight(rightKey) {
      return this.right = rightKey ? createNode(rightKey) : null;
    }
  };
}

function createBT(rootKey) {
  const root = createNode(rootKey);
  return {
    root,
    nodes: [],
    serialize(node) {
      if (node) {
        this.nodes.push(node.val);
        this.serialize(node.left);
        this.serialize(node.right);
      } else {
        this.nodes.push("#");
      }
      return this.nodes.join(" ");
    },
    deserialize(str) {
      
      function* getVals() {
        for (let val of str.split(" ")) {
          yield val;
        }
      }
      
      const helper = (pointer) => {
        let { value , done } = pointer.next();
        if (value === "#" || done) return null;

        let node = createNode(value);
        node.left = helper(pointer);
        node.right = helper(pointer);
        return node;
      };
      return helper(getVals());
    }
  };
}
////////Construct the tree///////////
const tree = createBT("root");
const root = tree.root;
const left = root.addLeft("left");
root.addRight("right");
left.addLeft("left.left");
left.addRight("left.right");

///////////Serialize and deserialize//////////////
const ser = tree.serialize(root); // root left left.left # # # right # #
const dec = tree.deserialize(ser).left.left.val; // left.left
console.log(ser, dec)