算法学习记录（二十五）

问：

1. 莫里斯遍历
2. 莫里斯遍历改先中后序遍历

解：

1. 流程：①判断当前是否有左节点，没有则当前节点指向右节点；②找到左树的最右节点，若最右节点的右节点为null，将其指向当前节点，当前节点指向左节点；若最右节点的右节点为当前节点，将其重新指向null，当前节点指向右节点

function Morris(root) {
    let curNode = root
    while (curNode) {
        if (curNode.left) {
            let mostRight = curNode.left
            while (mostRight.right && mostRight.right !== curNode) {
                mostRight = mostRight.right
            }
            if (mostRight.right === null) {
                mostRight.right = curNode
                curNode = curNode.left
                continue
            }
            if (mostRight.right === curNode) {
                mostRight.right = null
            }
        }
        curNode = curNode.right
    }
}

2. 先序：只到达一次的节点直接打印；到达两次的节点，第一次到达时打印

function Morris(root) {
    let curNode = root
    while (curNode) {
        if (curNode.left) {
            let mostRight = curNode.left
            while (mostRight.right && mostRight.right !== curNode) {
                mostRight = mostRight.right
            }
            if (mostRight.right === null) {
                console.log(curNode)
                mostRight.right = curNode
                curNode = curNode.left
                continue
            }
            if (mostRight.right === curNode) {
                mostRight.right = null
            }
        }
        console.log(curNode)
        curNode = curNode.right
    }
}

中序：只到达一次的节点直接打印；到达两次的节点，第二次到达时打印

function Morris(root) {
    let curNode = root
    while (curNode) {
        if (curNode.left) {
            let mostRight = curNode.left
            while (mostRight.right && mostRight.right !== curNode) {
                mostRight = mostRight.right
            }
            if (mostRight.right === null) {
                mostRight.right = curNode
                curNode = curNode.left
                continue
            }
            if (mostRight.right === curNode) {
                mostRight.right = null
            }
        }
        console.log(curNode)
        curNode = curNode.right
    }
}

后序：到达两次的节点，第二次到达时，逆序打印左树的右边界

function Morris(root) {
    let curNode = root
    while (curNode) {
        if (curNode.left) {
            let pre = null
            let mostRight = curNode.left
            while (mostRight.right && mostRight.right !== curNode) {
                mostRight = mostRight.right
            }
            if (mostRight.right === null) {
                mostRight.right = curNode
                curNode = curNode.left
                continue
            }
            if (mostRight.right === curNode) {
                mostRight.right = null
                printNode(curNode.left)
            }
        }
        console.log(curNode)
        curNode = curNode.right
    }
}
// 处理完再反转回去
function printNode(node) {
    const reverseNode = reverseTree(node)
    let curNode = reverseNode
    while (curNode) {
        console.log(curNode)
        curNode = curNode.right
    }
    reverseTree (reverseNode)
}
// 类似反转链表
function reverseTree(node) {
    let pre = null
    let temp = null
    while(node){
        temp = node.right
        node.right = pre
        pre = node
        node = temp
    }
    return pre
}