【算法】还原二叉树

一、根据前序中序还原二叉树

let qian = ["A", "B", "D", "E", "C", "F", "G"];
let zhong = ["D", "B", "E", "A", "F", "C", "G"];

function Node(value) {
	this.value = value;
	this.left = null;
	this.right = null;
}
function f1(qian, zhong) {
	if(qian == null || zhong == null || qian.length == 0 || zhong.length == 0 || qian.length != zhong.length) return null;
	let root = new Node(qian[0]);
	console.log(root.value); // ABDECFG
	let index = zhong.indexOf(root.value);
	let qianLeft = qian.slice(1, index + 1);
	let qianRight = qian.slice(index + 1, qian.length);
	let zhongLeft = zhong.slice(0, index);
	let zhongRight = zhong.slice(index + 1, zhong.length);
	root.left = f1(qianLeft, zhongLeft);
	root.right = f1(qianRight, zhongRight);
	return root;
}
const node = f1(qian, zhong);
console.log(node);
// Node {
// 	value: 'A',
// 	left: Node {
// 	  value: 'B',
// 	  left: Node { value: 'D', left: null, right: null },
// 	  right: Node { value: 'E', left: null, right: null }
// 	},
// 	right: Node {
// 	  value: 'C',
// 	  left: Node { value: 'F', left: null, right: null },
// 	  right: Node { value: 'G', left: null, right: null }
// 	}
// }

root.left = f1(qianLeft, zhongLeft);

root.right = f1(qianRight, zhongRight);

左边的树从左到右遍历完后，再遍历右边的树

右边的树在遍历过程中，也遵循从左到右遍历

都遍历完了最后从栈中拿出节点对应赋值还原出树

二、根据中序后序还原二叉树

let zhong = ["D", "B", "E", "A", "F", "C", "G"];
let hou = ["D", "E", "B", "F", "G", "C", "A"];

function Node(value) {
	this.value = value;
	this.left = null;
	this.right = null;
}
function f1(zhong, hou) {
	if(zhong == null || hou == null || zhong.length == 0 || hou.length == 0 || zhong.length != hou.length) return null;
	let root = new Node(hou[hou.length - 1]);
	let index = zhong.indexOf(root.value);
	let zhongLeft = zhong.slice(0, index);
	let zhongRight = zhong.slice(index + 1, zhong.length);
	let houLeft = hou.slice(0, index);
	let houRight = hou.slice(index, hou.length - 1);
	root.left = f1(zhongLeft, houLeft);
	root.right = f1(zhongRight, houRight);
	return root;
}
const node = f1(zhong, hou);