100. 相同的树🔖BFS🔖DFS

• https://leetcode-cn.com/problems/same-tree/

BFS

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {boolean}
 */
var isSameTree = function(p, q) {
    if (p === null && q === null) return true
    if (p === null || q === null) return false
    let que_1 = [p]
    let que_2 = [q]
    while(que_1.length !== 0) {
        let que_1_size = que_1.length
        let que_2_size = que_2.length
        if (que_1_size !== que_2_size) return false
        for (let i = 0; i<que_1_size; i++) {
            let mp1 = que_1.shift()
            let mp2 = que_2.shift()
            if (mp1.val !== mp2.val) return false
            if (mp1.left && mp2.left) {
                que_1.push(mp1.left)
                que_2.push(mp2.left)
            } else if(!mp1.left && !mp2.left){} else {
                return false
            }
            if (mp1.right && mp2.right) {
                que_1.push(mp1.right)
                que_2.push(mp2.right)
            } else if(!mp1.right && !mp2.right){} else {
                return false
            }
        }
    }
    return true
};

DFS

肝不动了，直接暴力run ok就行

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} p
 * @param {TreeNode} q
 * @return {boolean}
 */
var isSameTree = function(p, q) {
    let result = true
    if (p === null && q === null) return result
    if (p === null || q === null) return !result
    let arr_1 = []
    let arr_2 = []
    let dfs = (data, arr) => {
        if (data) {
            arr.push(data.val)
            if (data.left) {
                dfs(data.left, arr)
            } else {
                arr.push(99999)
            }
            if (data.right) {
                dfs(data.right, arr)
            } else {
                arr.push(99999)
            }
        } 
    }
    dfs(p, arr_1)
    dfs(q, arr_2)
    if (arr_1.length !== arr_2.length) result = false
    arr_1.forEach((data, index) => {
        if (data !== arr_2[index]) {
            result = false
        }
    })
    return result
};