二叉树的先中后序遍历

先序遍历： 根左右

1. 访问根节点
2. 对根节点的左子树进行先序遍历
3. 对根节点的右子树进行先序遍历

非递归版代码实现

const bt = {
    val: 50,
    left: {
      val: 10,
      left: {
        val: 5,
        left: null,
        right: null,
      },
      right: {
        val: 15,
        left: null,
        right: null,
      },
    },
    right: {
      val: 70,
      left: {
        val: 60,
        left: null,
        right: null,
      },
      right: {
        val: 80,
        left: null,
        right: null,
      },
    },
  };

const preorder = (root) => {
    if (!root) { return; }
    const stack = [root];
    while (stack.length) {
        const n = stack.pop();
        console.log(n.val);
        if (n.right) stack.push(n.right);
        if (n.left) stack.push(n.left);
    }
};

// 50 10 5 15 70 60 80 

preorder(bt);

中序遍历： 左根右

1. 对根节点的左子树进行中序遍历
2. 访问根节点
3. 对根节点的右子树进行中序遍历

非递归版代码实现

const bt = {
    val: 56,
    left: {
      val: 22,
      left: {
        val: 10,
        left: null,
        right: null,
      },
      right: {
        val: 30,
        left: null,
        right: null,
      },
    },
    right: {
      val: 81,
      left: {
        val: 77,
        left: null,
        right: null,
      },
      right: {
        val: 92,
        left: null,
        right: null,
      },
    },
  };

const inorder = (root) => {
    if (!root) { return; }
    const stack = [];
    let p = root;
    while (stack.length || p) {
        while (p) {
            stack.push(p);
            p = p.left;
        }
        const n = stack.pop();
        console.log(n.val);
        p = n.right;
    }
};

// 10  22 30 56  77 81 92

inorder(bt);

​

后序遍历： 左右根

1. 对根节点的左子树进行后序遍历
2. 对根节点的右子树进行后序遍历
3. 访问根节点

非递归版代码实现

const bt = {
    val: 23,
    left: {
      val: 16,
      left: {
        val: 3,
        left: null,
        right: null,
      },
      right: {
        val: 22,
        left: null,
        right: null,
      },
    },
    right: {
      val: 45,
      left: {
        val: 37,
        left: null,
        right: null,
      },
      right: {
        val: 99,
        left: null,
        right: null,
      },
    },
  };

const postorder = (root) => {
    if (!root) { return; }
    const outputStack = [];
    const stack = [root];
    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){
        const n = outputStack.pop();
        console.log(n.val);
    }
};

postorder(bt);