103 lines
2.7 KiB
JavaScript
103 lines
2.7 KiB
JavaScript
/**
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* Definition for a binary tree node.
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* function TreeNode(val, left, right) {
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* this.val = (val===undefined ? 0 : val)
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* this.left = (left===undefined ? null : left)
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* this.right = (right===undefined ? null : right)
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* }
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*/
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/**
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* @param {TreeNode} root
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* @return {void} Do not return anything, modify root in-place instead.
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*/
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import TreeNode from './tool';
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const flatten = function (root) {
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};
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/*
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思路：直接先序遍历整个二叉树，将整个节点按照顺序存进数组最后遍历它们，把它们连接起来
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*/
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function f1(root) {
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// 如果为空返回
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if (!root) return root;
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const preNodes = [];// 储存先序遍历的数组
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const dummyHead = new TreeNode(1); // 哑节点，用于构建链表
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// 先序遍历
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const stack = [root]; // 先序遍历需要用到的栈
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while (stack.length) {
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const node = stack.pop(); // 取出根节点
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preNodes.push(node);
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if (node.right) stack.push(node.right);
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if (node.left) stack.push(node.left);
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}
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// 将数组中所有的节点通过right链接到dummyHead
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let preNode = dummyHead;
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for (let i = 0; i < preNodes.length; i++) {
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preNodes[i].left = null; // 左节点滞空
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preNode.right = preNodes[i];
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preNode = preNodes[i];
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}
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return dummyHead.right;
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}
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/*
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利用栈来前序遍历，思路和上面迭代一致
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*/
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function f2(root) {
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const preorderTraversal = (node, nodeList) => {
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if (node) {
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nodeList.push(node);
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preorderTraversal(node.left, nodeList);
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preorderTraversal(node.right, nodeList);
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}
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};
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if (!root) return root;
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const list = [];
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// 递归遍历二叉树，把节点放进list中
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preorderTraversal(root, list);
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// 将所有节点链接起来
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const dummyHead = new TreeNode(-1); // 利用哑节点来处理头节点问题
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let cur = dummyHead;
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for (let i = 0; i < list.length; i++) {
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cur.right = list[i];
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cur.left = null;
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cur = list[i];
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}
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return dummyHead.right;
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}
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/*
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在前序遍历的同时设置节点，根据前序遍历的顺序，会先处理左子树，再处理右子树，所以只要先把右子树
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压入栈中，再把左子树压入栈中，这样当所有的左子树处理完毕，会继续处理右子树
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*/
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function f3(root) {
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if (!root) return root;
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const stack = [root];
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// 要链接到的节点，初始为空，表示字节的的前面一个节点
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let preNode = null;
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while (stack.length) {
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const node = stack.pop();
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// 将当前节点链接到preNode.right
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if (preNode) {
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preNode.right = node;
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preNode.left = null;
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}
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// 将右子树压入栈中
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if (node.right) stack.push(node.right);
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// 将左子树压入栈中
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if (node.left) stack.push(node.left);
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preNode = node;
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}
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}
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