二叉树的先中后序遍历-JS非递归实现

const bt = {
    val: 1,
    left: {
        val: 2,
        left: {
            val: 4,
            left: null,
            right: null,
        },
        right: {
            val: 5,
            left: null,
            right: null,
        },
    },
    right: {
        val: 3,
        left: {
            val: 6,
            left: null,
            right: null,
        },
        right: {
            val: 7,
            left: null,
            right: null,
        },
    },
};

//可以用堆栈来模拟递归

//先序遍历
const preorder = (root) => {
    if(!root) {return; }
    const stack = [root];  //创建一个栈
    while (stack.length) {
        const n = stack.pop();
        console.log(n.val);  
        if(n.right) stack.push(n.right); //根左右，栈是后进先出，所以先把右结点压栈
        if(n.left) stack.push(n.left);
    }
}
preorder(bt)

//中序遍历
const inorder = (root) => {
    if(!root) {return; }
    const stack = [];  //创建一个栈
    let p = root;
    while(stack.length || p) {
        while(p) {      
            stack.push(p); //左根右,遍历所以左结点压入栈中
            p = p.left;
        }
        const n = stack.pop();
        console.log(n.val); //弹出就是 左结点和中结点
        p = n.right; //把右结点压入，参与循环
    }
}

inorder(bt);

//后序遍历
const postorder = (root) => {    //后序遍历是左 右 根，先序遍历得到的是根 左 右
    if(!root) {return; }  // 把前序遍历的顺序调换一下得到 根 右 左，压栈再弹出就变成了 左 右  根
    const outputStack = [];
    const stack = [root];  //需要两个栈
    while (stack.length) {
        const n = stack.pop();
        outputStack.push(n);
        if(n.left) stack.push(n.left);
        if(n.right) stack.push(n.right);
    }
    while(outputStack.length) {  //倒叙输出即可
        const n = outputStack.pop();
        console.log(n.val);
    }
}

postorder(bt);